THE PRACTICE STUDIO
Go slowly. Ask why. Make it yours.
PRACTICE / Time of flight
Time back to launch height
A ball is launched with velocity components 12 m/s horizontally and 15 m/s upward.
When does it return to its launch height?
Organize the givens & assumptions
- ux = 12 m/s
- uy = 15 m/s
- g = 10 m/s²
Uniform g = 10 m/s²; no drag; launch and landing heights equal.
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.
Opening this prompt marks this practice attempt as assisted.
Problem: Time back to launch height A ball is launched with velocity components 12 m/s horizontally and 15 m/s upward. Question: When does it return to its launch height? Givens: - ux = 12 m/s - uy = 15 m/s - g = 10 m/s² Assumptions: Uniform g = 10 m/s²; no drag; launch and landing heights equal. Choices (LaTeX): A. 0\ \mathrm s B. 3\ \mathrm s C. 1.5\ \mathrm s D. 2.4\ \mathrm s 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.
Start with the smallest hint.
Work through it with me.
One decision at a time. Pause, predict, then unpack the algebra.
Set the vertical displacement to zero and choose the nonzero root, not the launch instant.
Choose a physical relation
Returning to launch height sets vertical displacement to zero; the nonzero time root gives the flight duration.
Isolate what is asked
Set the vertical displacement to zero and choose the nonzero root, not the launch instant.
Why is this step valid?
Factor out t to separate the launch instant from the later return, leaving T = 2uy/g.
Substitute and interpret
Flight lasts 3 s; horizontal speed does not enter the vertical return condition.
The horizontal velocity affects range but cannot alter the equal-height flight time.
TRY THE IDEA AGAIN
If the landing point lies below launch, can 2uy/g still give the landing time?
YOUR LEARNING NOTE
How does this idea feel now?
Confidence is a reflection, not proof of mastery.
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