Q 7.     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?

Answers (1)
H Harsh Kankaria

Considering downward direction as positive direction.

Given, Height from which ball is dropped, s = 20\ m

Acceleration of the ball, a = 10\ ms^{-2}

Initial velocity, u = 0\ ms^{-1}

(i) We know, v^2 = u^2 + 2as

\\ \implies v^2 = 0^2 + 2(10)(20) \\ \implies v^2 = 400 \\ \implies v = 20\ ms^{-1}(In downward direction)

Therefore, the ball will strike the ground with a velocity of 20\ ms^{-1}

(ii) Now, we know, v = u + at

\\ \implies 20 = 0 + 10t \\ \implies t = 2\ s

Therefore, the ball reaches the ground in 2\ s.

Note: v = -20\ ms^{-1} was rejected because in this case, the negative sign implies the velocity in upward direction, which is opposite to the direction of the motion of the ball(before collision).