A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h/3 in both the directions.

Solution

Physics sir · 4 steps

Stand at the top and look down: the ball reaches h/3 a time τ before the top and a time τ after it.

1. Time to the top. Going up to h is the reverse of falling h from rest. So the ball reaches the top at T = √(2h/g).
2. Distance below top. The point h/3 is h - h/3 = 2h/3 below the top. Falling that far from rest takes τ = √(2/g × 2h/3) = √(4h/3g).
3. Use symmetry. The ball passes h/3 on the way up at t₁ = T - τ and on the way down at t₂ = T + τ.

   t₁/t₂ = (√2 - √(4/3))/(√2 + √(4/3)) = (√3 - √2)/(√3 + √2)

4. So. Cancel the common factor √(h/g), then multiply the top and bottom by √(3/2). The ratio is (√3 - √2)/(√3 + √2), option (c).

The idea to keep. For a vertical throw, both crossing times of a height are (time to top) minus and plus (time to fall from the top to that height). Always measure that fall from the top, not from the ground.

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