From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.

Solution

Physics sir · 4 steps

Write the tower height twice, once for each ball, and the unknown speed drops out of the problem.

1. Set up both throws. Take downward as positive, with tower height h and launch speed u. The upward ball starts with velocity -u and takes t₁ = 6 s. The downward ball starts with +u and takes t₂ = 1.5 s.

   h = -u t₁ + 1/2 g t₁² = u t₂ + 1/2 g t₂²

2. Find u, then h. Equating the two expressions gives u(t₁ + t₂) = 1/2 g (t₁² - t₂²), so u = 1/2 g (t₁ - t₂). Putting this back gives h = 1/2 g t₁ t₂. With g = 10 m s⁻², u = 22.5 m s⁻¹ and h = 45 m.
3. Drop from rest. For the third ball, h = 1/2 g t₃². Comparing with h = 1/2 g t₁ t₂ gives t₃ = √(t₁ t₂) = √(6 × 1.5) = √9.
4. So. The ball released from rest reaches the ground in 3 s.

The common slip: 3.75. Averaging the two times as (6 + 1.5)/2 gives 3.75, but the drop time is the geometric mean √(6 × 1.5), not the arithmetic mean.

The idea to keep. Up-throw time t₁ and down-throw time t₂ at the same speed give the free-drop time √(t₁ t₂). It is the geometric mean, and it does not depend on g or u.

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