Question

9. A ball is released from a height and it reaches the
ground in 3 s. If g = 9.8 m s?, find :
(a) the height from which the ball was released,
(b) the velocity with which the ball will strike the
ground. ​

Answer 1

Explanation:

$u = 0 \\ t = 3 \\ g = 9.8 \\ s = ut + \frac{1}{2} a {t}^{2} \\ s = 4.9 \times 9 \\ s = 44.1 \\ {v}^{2} - {u}^{2} = 2as \\ {v}^{2} = 19.6 \times 44.1 \\ v = \sqrt{44.1 \times 19.6}$

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