The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is :

Solution

Physics sir · 4 steps

Same man, same ball, so the same launch speed: find u²/g from the vertical throw and reuse it.

1. Vertical throw. Thrown straight up, the ball stops at the top, so 0 = u² - 2gH. This gives H = u²/2g, the greatest height for launch speed u.
2. Find the speed. With H = 136 m, we get u²/2g = 136, so u²/g = 272 m.
3. Maximum range. On flat ground the range is R = (u² sin 2θ)/g. It is largest when sin 2θ = 1, that is at θ = 45°.

   Rₘₐₓ = u²/g = 2H

4. So. Rₘₐₓ = 2 × 136 = 272 m, which is option (b).

The idea to keep. For the same launch speed, the farthest horizontal throw u²/g is exactly twice the highest vertical throw u²/2g. Fix u from one, then use it in the other.

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