Question

A ball is gently dropped from a height of 20m. If its velocity increases uniformly at the rate of 10 ms^-2 . With what velocity will it strike the ground? After what time will it strike the ground?

Answer 1

Given: 

height = 20 m 

acceleration = 10 ms 

Formula: 

For free falling body,

1. v = gt 

2. s = 1/2 gt²

time taken to strike the ground 

s = 1/2 gt² 

20m = 1/2 × 10 ms^-2 × t²

t² = \(\frac{40 m}{10 ms^{-2}}\)= 4s² 

∴ t = 2s 

3. Velocity of the ball when it strikes ground v = gt

v = 10ms^-2 × 2s 

v = 20ms^-1