# 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?

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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