A body projected vertically upwards with a certain speed from the top of a tower reaches the ground in t₁. If it is projected vertically downwards from the same point with the same speed, it reaches the ground in t₂. Time required to reach the ground, if it is dropped from the top of the tower, is :

Solution

Physics sir · 4 steps

Write the same tower height twice, once for each throw, and eliminate the unknown speed u.

1. Set up signs. Take downward as positive, with the origin at the top of the tower. Let the tower height be h and the launch speed be u. Both throws then land at displacement +h.
2. Two equations. Thrown up, the initial velocity is -u: h = -u t₁ + 1/2 g t₁². Thrown down, it is +u: h = u t₂ + 1/2 g t₂².
3. Eliminate u. Multiply the first equation by t₂ and the second by t₁, then add. The u terms cancel, and the common factor (t₁ + t₂) divides out.

   h(t₁ + t₂) = 1/2 g t₁ t₂ (t₁ + t₂) ⇒ h = 1/2 g t₁ t₂

4. So. Dropped from rest, h = 1/2 g t². Comparing with h = 1/2 g t₁ t₂ gives t² = t₁ t₂, so t = √(t₁ t₂), option (b).

The idea to keep. When the same height comes from two throws, write h for each with a sign convention and eliminate the unknown speed. Then check that the answer has the dimension of time.

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