A ball is thrown up vertically with a certain velocity so that, it reaches a maximum height h. Find the ratio of the times in which it is at height h/3 while going up and coming down respectively.

Solution

Physics sir · 3 steps

Measure time from the top: the ball takes the same time to fall to h/3 as to rise from it.

1. Use the top. Let the ball take time T to reach the top, so h = 1/2gT². The motion is symmetric about the top, so it passes h/3 a time t' before the top and again a time t' after it.
2. Find t'. From the top the ball starts at rest, and h/3 lies 2h/3 below the top. Free fall through 2h/3 takes t', with 2h/3 = 1/2gt'². Dividing by h = 1/2gT² gives t' = T√(2/3).

   t₁/t₂ = (T - t')/(T + t') = (1 - √(2/3))/(1 + √(2/3)) = (√3-√2)/(√3+√2)

3. So. Going up it is at h/3 at time T - t', coming down at time T + t'. Their ratio is (√3-√2)/(√3+√2), option (b).

The idea to keep. For up-and-down motion under gravity, measure time from the highest point. There the ball is at rest, so the two crossing times are just T ∓ t'.

Answer from the NTA final answer key. Solution by PrepWiser.