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

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance traveled by the ball (s) = 20m

Acceleration (a) = 10 ms^-2

By third motion equation, \( v^2 - u^2 = 2 a s \)

Therefore,

= 2×(10)×(20) + 0

v² = 400

Therefore, v= 20m/s

The ball hits the ground with a velocity of 20 meters per second.

By first motion equation, \( v = u + at \)

Therefore, \( 20 = 0 + 10 t \)

\(\Rightarrow \) t = 2 seconds

Therefore, the ball reaches the ground after 2 seconds.