MixedJEE Physics · Original learning card10 original chapter questions

Time of Flight and Vertical Timing

The flight time is obtained from y_f - y_0 = u_y T - one-half gT squared. When y_f = y_0, the nonzero root is T = 2u sin theta/g.

Why this shows up in the exam

Scheduling interception at a height · Comparing projectiles with different angles · Finding launch components from ascent time

Learn the idea

Flight timing is controlled entirely by the vertical component and the chosen final height. Horizontal motion decides where the projectile lands, but vertical motion decides when it reaches a level. Equal launch and landing heights give a symmetric up-and-down time.

🧠 Memory hook: Time comes from vertical motion; equal levels make ascent and descent times match.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Delta y = u_y T - (1/2)gT² — general vertical timing equation
  • T = 2u sin(theta)/g — flight time for equal launch and landing levels
  • t_up = u sin(theta)/g — time to the highest point

How to approach it

  1. 1Write the vertical displacement with signs
  2. 2Solve the time equation and reject nonphysical roots
  3. 3Use horizontal motion only after the time is known

Common slip-ups that cost marks

  • •Using equal-level time for an elevated landing
  • •Taking the zero root as the flight time
  • •Using horizontal speed in the vertical timing equation

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?

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