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)

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).

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