A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of 100 m from the foot of the tower. A body of mass 2 M thrown at a velocity v/2 from the top of the tower of height 4 H will touch the ground at a distance of _______ m.

Solution

Physics sir · 4 steps

Range is speed times fall time, and fall time grows as root of height. Mass plays no part at all.

1. Fall time. Thrown horizontally, the body starts with zero vertical velocity, so H = 1/2gt² gives t = √(2H/g). Gravity gives every mass the same acceleration, so M and 2M do not matter.

   R = v t = v√(2H/g)

2. Compare the cases. Height becomes 4H, so the fall time becomes √4 = 2 times longer. Speed becomes v/2, half as much.
3. Take the ratio. R₂/R₁ = 1/2 × 2 = 1, so R₂ = R₁ = 100 m.
4. So. The second body lands 100 m from the foot of the tower.

The common slip: 200. Treating the fall time as proportional to H instead of √H makes the time 4 times longer and gives 200 m, not 100 m.

The idea to keep. For a horizontal throw, R ∝ v√H. Fall time depends only on height, horizontal speed only stretches the range, and mass never enters.

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