A boy throws a ball into air at 45° from the horizontal to land it on a roof of a building of height H. If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of H is ____ m.
(g=10 m/s²)

Solution

Physics sir · 4 steps

Forget the horizontal motion: the time to the top gives the vertical speed, and that gives the height.

1. Vertical launch speed. At the highest point the vertical velocity is zero. So u_y = g tₜₒₚ = 10 × 2 = 20 m s⁻¹. The 45° angle is not needed for the height.
2. Height at the top. The maximum height is u_y²/2g = 400/20 = 20 m, reached at t = 2 s.
3. Fall for one second. From the top, the ball starts with zero vertical speed and falls for 3 - 2 = 1 s. It drops 1/2 g t² = 1/2 × 10 × 1² = 5 m.

   H = u_y t - 1/2 g t² = 20 × 3 - 1/2 × 10 × 3² = 60 - 45 = 15 m

4. So. The roof is 20 - 5 = 15 m high, which matches the direct formula. The answer is (d).

The idea to keep. In projectile questions about height and time, use only the vertical motion. The time to reach the top gives u_y = g t, and every height follows from that.

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