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PRACTICE / Projectile motion

What happens at the highest point?

A ball is launched from level ground with horizontal velocity 12 m/s and upward vertical velocity 16 m/s. Ignore air resistance and take g = 10 m/s².

Which pair gives its speed and acceleration magnitude at the highest point?

Organize the givens & assumptions

Horizontal launch velocity: +12 m/s
Vertical launch velocity: +16 m/s
Gravity: 10 m/s² downward

Point particle, negligible air resistance, constant gravity and an inertial ground frame. Right and up are positive. The ball is in free flight, before landing.

Choose one answerA

0\ \mathrm{m\,s^{-1}},\quad 0\ \mathrm{m\,s^{-2}}

12\ \mathrm{m\,s^{-1}},\quad 0\ \mathrm{m\,s^{-2}}

12\ \mathrm{m\,s^{-1}},\quad 10\ \mathrm{m\,s^{-2}}

20\ \mathrm{m\,s^{-1}},\quad 10\ \mathrm{m\,s^{-2}}

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Problem: What happens at the highest point?
A ball is launched from level ground with horizontal velocity 12 m/s and upward vertical velocity 16 m/s. Ignore air resistance and take g = 10 m/s².
Question: Which pair gives its speed and acceleration magnitude at the highest point?
Givens:
- Horizontal launch velocity: +12 m/s
- Vertical launch velocity: +16 m/s
- Gravity: 10 m/s² downward
Assumptions: Point particle, negligible air resistance, constant gravity and an inertial ground frame. Right and up are positive. The ball is in free flight, before landing.
Choices (LaTeX):
A. 0\ \mathrm{m\,s^{-1}},\quad 0\ \mathrm{m\,s^{-2}}
B. 12\ \mathrm{m\,s^{-1}},\quad 0\ \mathrm{m\,s^{-2}}
C. 12\ \mathrm{m\,s^{-1}},\quad 10\ \mathrm{m\,s^{-2}}
D. 20\ \mathrm{m\,s^{-1}},\quad 10\ \mathrm{m\,s^{-2}}

Ask for my approach first. Help me distinguish a physical misconception from an algebra error without revealing which choice is correct.

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.

A NUDGE, NOT A SPOILER

Start with the smallest hint.

YOUR PACE

Work through it with me.

One decision at a time. Pause, predict, then unpack the algebra.

THE REASONING

The ball stops rising. It doesn’t stop moving.

Choose the axes

Take right as positive x and up as positive y. With air resistance neglected, gravity is the only force during flight.

a_x=0,\qquad a_y=-10\ \mathrm{m\,s^{-2}}

Why is this step valid?

Acceleration is determined by the net force. The ball's instantaneous velocity does not determine whether gravity acts.

Find the top of the flight

The ball stops rising when its vertical velocity reaches zero. Solve for the time at that instant.

\begin{aligned}v_y&=16-10t=0\\ t&=1.6\ \mathrm{s}\end{aligned}

Why is this step valid?

The vertical velocity changes sign from positive to negative at the maximum height. Its rate of change is still −10 m/s².

Keep the horizontal motion

There is no horizontal acceleration, so the horizontal velocity remains 12 m/s. Speed is the magnitude of the full velocity vector.

v=\sqrt{12^2+0^2}=12\ \mathrm{m\,s^{-1}}

Why is this step valid?

Speed is zero only when every velocity component is zero. A zero vertical component alone is not enough.

Check the acceleration

At the top, gravity still accelerates the ball downward. The speed and acceleration magnitude are therefore option C.

\begin{aligned}|\mathbf{a}|&=\sqrt{0^2+(-10)^2}\\ &=10\ \mathrm{m\,s^{-2}}\end{aligned}

Why is this step valid?

A moment later the ball has downward vertical velocity because the acceleration was already downward at the top.

Check it another way: energy

The maximum height is uy²/(2g) = 12.8 m. The launch speed is √(12² + 16²) = 20 m/s. Energy conservation gives v² = 144 m²/s², so the speed is 12 m/s.

\begin{aligned}v^2&=u^2-2gh\\ &=20^2-2(10)(12.8)\\ &=144\ \mathrm{m^2\,s^{-2}}\end{aligned}

TRY THE IDEA AGAIN

If the launch were perfectly vertical, which pair describes speed and acceleration magnitude at the top?

YOUR LEARNING NOTE

How does this idea feel now?

Confidence is a reflection, not proof of mastery.

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THE MOTION LAB

Make the idea move.

Interactive

Scrub through the flight. Watch what changes, and what stays the same.

Velocity componentsGravity

Flight time 0.00 s

Playback pace

HORIZONTAL vₓ12.00 m/s

VERTICAL vᵧ16.00 m/s

ACCELERATION aᵧ−10.00 m/s²

What if the launch angle changed?

Exploration keeps launch speed at 20 m/s and gravity at 10 m/s². The question on the left keeps its original values.

Launch angle 53.1°

Using the question’s original launch.

Level ground · No air resistance · Constant gravity.
Arrow scales differ for velocity and acceleration. Motion plays slowly; use the clock for physical time. With reduced motion enabled, Play advances one step.

WHAT YOU’RE PRACTISING

Compare cases

Projectile motion Velocity and acceleration

A USEFUL STUDY LOOP

State the knowns and the unknown.
Choose a model and test its assumptions.
Try the smallest hint you need.
Explain each step in your own words.
Test the idea with a changed condition.

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