When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. It reaches the bottom of the lake 4.0 s after it is dropped. The approximate depth of the lake is :

Solution

Physics sir · 4 steps

Split the 4 s: find the time in air first, and the rest is spent sinking at steady speed.

1. Time in air. The ball is dropped from rest, so h = 1/2gt₁². With h = 4.9 m and g = 9.8 m s⁻², this gives 4.9 = 4.9 t₁², so t₁ = 1 s.
2. Speed at water. It hits the water with v = gt₁ = 9.8 × 1 = 9.8 m s⁻¹.
3. Time in water. The whole trip takes 4.0 s, so the ball sinks for t₂ = 4.0 - 1 = 3 s. In the water its speed stays constant at v, so there is no acceleration term.

   d = v t₂ = 9.8 m s⁻¹ × 3 s = 29.4 m

4. So. The lake is about 29.4 m deep, which is option (b).

The idea to keep. When the motion changes type partway, break it into stages. Each stage hands its final speed and its time to the next one.

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