A ball at the top of its flight seems to pause. Look more closely: it has stopped rising, but it may still be travelling sideways. Following one clock through two directions lets us explain that moment without giving gravity a tea break.
We first build projectile motion from component equations. Then we change viewpoint and examine relative motion, where a complicated pair of paths can become a simple separation problem. Every example keeps the frame, clock and physical limits explicit.
What you will learn to do
- Derive flight properties from horizontal and vertical equations sharing one time.
- Use the correct landing height and check obstacle events.
- Explain why velocity can be horizontal while acceleration remains downward.
- Subtract velocity vectors in a common frame.
- Distinguish minimum-time crossing, drift cancellation, collision and closest approach.
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BEFORE YOU READ
Bring back one idea
Try a sentence or a sketch before opening the comparison. If this is new, use the explanation to get started.
What does zero vertical velocity tell you about horizontal velocity?
Compare your reasoning
Nothing by itself. Perpendicular components are separate quantities. In ideal projectile motion without air resistance, horizontal velocity stays constant while the vertical component changes under gravity.
Revisit the explanation →01 / Projectile motion and moving observers
Two components share one clock
A projectile is an object moving after launch under the forces included in the model. Our basic model includes only uniform downward gravity: air resistance, buoyancy, thrust, Earth curvature and rotation are neglected. Choose a horizontal x axis and an upward y axis in a frame treated as inertial. Then aₓ = 0 and aᵧ = −g throughout the flight.
The two component equations can be solved separately, but they describe the same object at the same time. You cannot use one flight time for the horizontal displacement and another for the vertical displacement. The common time is the link that reconstructs the curved trajectory.
The launch components are uₓ = u cos θ and uᵧ = u sin θ when θ is measured above +x. A horizontal launch has uᵧ = 0, not aᵧ = 0. A vertical launch has uₓ = 0, so it is best treated directly as one-dimensional motion rather than dividing by uₓ to obtain a trajectory equation.
Each expression follows from constant-acceleration motion along one axis. The same elapsed time t appears in both.
WORKED EXAMPLE
Find the flight time from the component that fixes it
A small ball leaves a horizontal ledge 20 m above a level floor at 6 m/s. Neglect air resistance and use g = 10 m/s². Find its landing time and horizontal travel.
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STOP & THINK
Immediately after a horizontal launch, which vertical quantities are zero?
- Both vertical velocity and vertical acceleration
- Vertical velocity only
- Vertical acceleration only
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: Two components share one clock Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): x=x_0+u_xt,\qquad y=y_0+u_yt-\frac12gt^2,\qquad v_x=u_x,\quad v_y=u_y-gt Meaning: Each expression follows from constant-acceleration motion along one axis. The same elapsed time t appears in both. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
02 / Projectile motion and moving observers
The highest point is an event, not a force balance
If uᵧ is positive, the vertical velocity decreases until it reaches zero. That instant is the apex. It is a turning point for the vertical coordinate, not necessarily a stop of the full motion: the horizontal velocity remains uₓ.
Set vᵧ = 0 to find ttop = uᵧ/g, then substitute into the vertical displacement equation. This gives the rise H = uᵧ²/(2g) above the launch point. The result depends on the vertical launch component, not directly on the full speed.
At the apex the acceleration is still downward with magnitude g. If uₓ is nonzero, velocity and acceleration happen to be perpendicular there. Just before the apex gravity reduces the speed; just after it gravity increases the speed. Neither observation implies a momentary disappearance of the force.
WORKED EXAMPLE
A horizontal velocity survives the highest point
A projectile starts with components uₓ = 12 m/s and uᵧ = 16 m/s. Take g = 10 m/s². Find the apex time, rise and speed there.
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Read the two velocity components at the same time
- Horizontal
- 12
- Vertical
- 16
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: The highest point is an event, not a force balance Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
03 / Projectile motion and moving observers
The landing surface decides which time is relevant
When launch and landing heights are equal, the vertical displacement equation factors into t(uᵧ − gt/2) = 0. One root is the launch event; the positive nonzero root is the return time 2uᵧ/g. Multiplying by uₓ gives the familiar range u² sin 2θ/g.
This range formula assumes equal endpoint heights, constant g and negligible air resistance. For fixed launch speed and a level landing surface, sin 2θ is largest at θ = 45°. Complementary angles then give the same range but different flight times and maximum heights. These conclusions are not universal optimisation rules for a cliff, slope or resisting medium.
For an unequal landing height, return to the component equations. Solve ylanding − y₀ = uᵧt − gt²/2, retain a physically relevant time and check whether an obstacle would intercept the trajectory earlier. A quadratic root is only a candidate event until the geometry and time domain are checked.
WORKED EXAMPLE
Landing below the launch point
A projectile starts 15 m above a horizontal floor with components uₓ = 8 m/s and uᵧ = 10 m/s. Use g = 10 m/s². Find the landing time, horizontal distance and velocity components just before impact.
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STOP & THINK
A projectile launches with uₓ = 9 m/s and uᵧ = 15 m/s and lands at its launch height. With g = 10 m/s², what is its horizontal range?
Answer in m.
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: The landing surface decides which time is relevant Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
04 / Projectile motion and moving observers
Eliminate time only when it helps the geometry
For nonzero uₓ, the horizontal equation gives t = (x − x₀)/uₓ. Substitute it into the vertical equation to obtain a parabola. The curvature coefficient is negative for upward y because gravity bends the trajectory downward.
The trajectory equation is useful for asking whether a projectile passes above a wall or through a specified point. It hides the timing, however. Two objects can pass through the same point at different times without colliding. If a moving target is involved, retain time until both positions are compared simultaneously.
A trajectory slope is dy/dx = vᵧ/vₓ when vₓ is nonzero. It describes the velocity direction, not the acceleration direction. Differentiating the parabola returns this changing slope and is a useful check that the component equations and path agree.
This form assumes a nonzero horizontal launch component. Its two terms have units of length.
WORKED EXAMPLE
Check the path at a wall before calling the ground the landing point
A projectile starts at ground level with uₓ = 10 m/s and uᵧ = 15 m/s. A wall 12 m high lies 10 m horizontally ahead. Use g = 10 m/s². Does the projectile clear the wall?
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers
Focus: Eliminate time only when it helps the geometry
Learning goals:
- Derive flight properties from horizontal and vertical equations sharing one time.
- Use the correct landing height and check obstacle events.
- Explain why velocity can be horizontal while acceleration remains downward.
- Subtract velocity vectors in a common frame.
- Distinguish minimum-time crossing, drift cancellation, collision and closest approach.
Equations in this section (LaTeX):
y-y_0=\frac{u_y}{u_x}(x-x_0)-\frac{g(x-x_0)^2}{2u_x^2}
Meaning: This form assumes a nonzero horizontal launch component. Its two terms have units of length.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.05 / Projectile motion and moving observers
Relative motion begins with a separation vector
The position of A relative to B is rA/B = rA − rB when both positions are measured in the same frame. This vector points from B to A. Differentiating gives vA/B = vA − vB; differentiating again gives aA/B = aA − aB. The order in the labels matches the subtraction order.
These are vector subtractions. Subtracting speed magnitudes works only in special collinear situations with carefully assigned signs. Two objects moving at the same speed in perpendicular directions have a nonzero relative velocity. They do not stand still relative to each other.
For classical motion, two observers whose axes remain parallel and whose relative velocity is constant agree on acceleration but generally disagree on velocity. Both can use Newton’s laws in inertial frames. An accelerating or rotating observer requires additional care; simple subtraction remains useful for parallel translating axes, but the force equation in the accelerated frame must include the frame acceleration.
All vectors on the right must use the same basis. Reversing A and B reverses the relative vectors.
WORKED EXAMPLE
Equal speeds need not mean zero relative speed
One cart moves east at 4 m/s while another moves north at 3 m/s. Find the first cart’s velocity relative to the second. Use east as +x and north as +y.
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STOP & THINK
If two particles have identical acceleration vectors, what follows?
- Their relative acceleration is zero
- Their relative velocity is zero
- Their separation is zero
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Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers
Focus: Relative motion begins with a separation vector
Learning goals:
- Derive flight properties from horizontal and vertical equations sharing one time.
- Use the correct landing height and check obstacle events.
- Explain why velocity can be horizontal while acceleration remains downward.
- Subtract velocity vectors in a common frame.
- Distinguish minimum-time crossing, drift cancellation, collision and closest approach.
Equations in this section (LaTeX):
\mathbf r_{A/B}=\mathbf r_A-\mathbf r_B,\quad\mathbf v_{A/B}=\mathbf v_A-\mathbf v_B,\quad\mathbf a_{A/B}=\mathbf a_A-\mathbf a_B
Meaning: All vectors on the right must use the same basis. Reversing A and B reverses the relative vectors.
Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding.
Tutoring approach:
Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.06 / Projectile motion and moving observers
A velocity relative to water is not a velocity relative to shore
For a boat in a steady current, add its velocity relative to the water to the water’s velocity relative to shore. The motor setting or stated swimming speed usually supplies the first quantity. The destination and crossing time are measured relative to shore. Keep the labels until the relationship is unambiguous.
To arrive directly opposite a starting point, the upstream component of the boat’s velocity relative to water must cancel the current. Only the remaining cross-river component carries it across. A boat aimed straight across instead uses its full water-relative speed for crossing and drifts downstream.
These are different optimisation questions. With fixed water-relative speed and uniform current parallel to the banks, aiming straight across minimises crossing time if any landing point is allowed. Aiming upstream to cancel drift minimises downstream displacement to zero when the boat speed exceeds the current. It usually increases crossing time.
WORKED EXAMPLE
Choose the heading for the stated destination
A river is 60 m wide and flows east at 3 m/s. A boat can move at 5 m/s relative to the water. Find its crossing time if it must land directly north of its starting point.
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STOP & THINK
A swimmer aims straight across a 40 m river at 2 m/s relative to the water. The current is 0.5 m/s parallel to the banks. How far downstream does the swimmer drift?
Answer in m.
MEMBERSAnswer checking, feedback and the explanation are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: A velocity relative to water is not a velocity relative to shore Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
07 / Projectile motion and moving observers
Meeting requires every position component to agree at one time
For constant relative velocity, separation evolves as rA/B(t) = rA/B(0) + vA/B t. A collision requires this vector to become zero: every component must vanish at the same physical time. Setting only one component to zero may merely place the objects on the same vertical or horizontal line.
For two projectiles acted on by the same uniform gravity, their relative acceleration is zero after both are in free flight. Subtracting their position equations cancels the gravity terms. Their relative path is therefore a straight line even though each ground-frame path is curved. This can simplify a meeting calculation considerably.
The cancellation only applies over the common interval with identical accelerations. A delayed launch, an object still supported by a ledge or significant unequal drag changes the model. First establish when each object is actually free, then compare positions and check that neither has already hit a surface.
WORKED EXAMPLE
Cancel common gravity without forgetting the ground
Two small balls are launched simultaneously from a height of 50 m. Ball A starts at x = 0 with velocity (10, 5) m/s; ball B starts at x = 20 m with velocity (0, 5) m/s. Use g = 10 m/s² and neglect their sizes and drag. Do their paths meet before reaching the ground?
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: Meeting requires every position component to agree at one time Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
08 / Projectile motion and moving observers
Closest approach need not be a collision
Two objects can come closest while still remaining separated. For a constant relative velocity, minimise the squared separation rather than taking a square root. Differentiating |r₀ + vt|² gives the condition (r₀ + vt)·v = 0: at closest approach, the separation vector is perpendicular to the relative velocity.
The unconstrained closest time is −r₀·v/|v|² when the relative velocity is nonzero. If the problem allows only t ≥ 0 and that result is negative, the objects are already moving apart and the closest future position is at t = 0. If the velocities change or a collision happens first, the constant-velocity calculation must stop at that event.
This method is useful for distinguishing a crossing of projected coordinates from an actual meeting. It also provides a geometric check: draw the straight relative path and drop the perpendicular from the origin. Its distance is the minimum separation when the perpendicular foot lies within the allowed time interval.
WORKED EXAMPLE
Same x coordinate, different y coordinate
An object has initial position (30, 10) m relative to an observer and a constant relative velocity (−5, 0) m/s. Find the closest future approach.
MEMBERSThe approach, step-by-step working and check are for members.Unlock →
Study with a tutor prompt
Copy this into your AI assistant, then describe where you got stuck. It asks the tutor to coach you one step at a time. Check its reasoning against the lesson.
Topic: Projectile motion and moving observers Focus: Closest approach need not be a collision Learning goals: - Derive flight properties from horizontal and vertical equations sharing one time. - Use the correct landing height and check obstacle events. - Explain why velocity can be horizontal while acceleration remains downward. - Subtract velocity vectors in a common frame. - Distinguish minimum-time crossing, drift cancellation, collision and closest approach. Equations in this section (LaTeX): Build the physical reasoning before choosing an equation. Start by checking the prerequisite idea I need most. Use a simple example, then change one condition to test my understanding. Tutoring approach: Act as a patient physics tutor. Ask what I understand and what I have tried, then wait for my reply. Teach one step at a time; do not give the final answer or a complete solution immediately. Start with a physical picture. Define unfamiliar symbols and operators, explain signs and units, and state the model's assumptions. Give progressively stronger hints only when I need them. Ask me to predict a change before calculating it. When I finish, help me check dimensions and a limiting case, then ask one related transfer question. If information is missing or ambiguous, ask rather than inventing it. Distinguish exact results from approximations and acknowledge uncertainty.
BRING THE IDEAS TOGETHER
What should stay with you
- A projectile has constant horizontal velocity only when the horizontal net force is zero.
- The apex condition is vᵧ = 0, not acceleration = 0.
- The usual range formula needs equal launch and landing heights.
- Relative velocity is a vector difference, not generally a difference of speeds.
- A meeting requires simultaneous agreement of all position components.
RETRIEVE, THEN EXPLAIN
Try it without the examples
Work these out before checking. If an answer surprises you, identify which assumption or step needs another look. A correct choice is most useful when you can explain why the alternatives fail.
SELF CHECK 1
A projectile has initial vertical component 18 m/s. With g = 10 m/s², after how many seconds does it reach its apex?
Answer in s.
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SELF CHECK 2
Two vehicles travel along the same line: A at +7 m/s and B at −3 m/s. What is A’s signed velocity relative to B?
Answer in m/s.
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SELF CHECK 3
Which change prevents direct use of R = u² sin(2θ)/g?
- Changing the mass while drag remains negligible
- Landing at a different height from launch
- Using a different horizontal origin
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SELF CHECK 4
Two objects pass through the same point at different times. What can you conclude?
- They collide
- Their paths intersect, but this does not establish a collision
- Their velocities must be identical
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APPLY THE CHAPTER
Take the ideas into a problem
Try a complete problem next. Commit to a method before revealing a hint, then compare your reasoning with the worked solution.
What happens at the highest point?Beginner · Free sample · Hints and guided reasoningTime back to launch heightIntermediate · Free sample · Hints and guided reasoningWhen do the two carts meet?Beginner · Free sample · Hints and guided reasoningCross the river, then measure the drift.Intermediate · Hints and guided reasoningA draining jet also fallsAdvanced · Hints and guided reasoningA horizontal launch from a ledgeIntermediate · Hints and guided reasoningHow high does the vertical part climb?Intermediate · Hints and guided reasoningThe two components share one clockIntermediate · Hints and guided reasoningA faster runner closes the gapIntermediate · Hints and guided reasoningCrossing without being swept downstreamIntermediate · Hints and guided reasoningRain seen from a moving trolleyIntermediate · Hints and guided reasoningCOME BACK TO THE IDEA
What can you explain later?
Try these from memory at your next study session, then again a few days later. Say your reasoning aloud or sketch it before looking. Adjust the gap to what you can explain.
Next study session
Compare a vertical throw and an angled throw at their highest points. Which can have zero speed, and what about acceleration?
Compare your reasoning
A purely vertical throw has zero speed at its peak. An angled throw retains horizontal speed. Both have downward acceleration g in the constant-gravity, drag-free model.
Revisit the explanation →A few days later: change the situation
A boat points straight across a river while the water flows downstream. Is its ground velocity in the pointing direction?
Compare your reasoning
Not generally. Add its velocity relative to the water to the water’s velocity relative to the bank. The shared idea is vector addition; the river current is not an extra acceleration by itself.
Revisit the explanation →Compare the physical reason, the assumptions and the conclusion. Your wording can differ. If one part is missing, revisit that section and try explaining it again after a break.
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