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A ball is released from a height and it reaches the ground in 3s. If g=9.8m/s^2 , find:

(a) The height from which the ball was released,

(b) The velocity with which the ball will strike the ground. Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given: t=3s, g = 9.8m/s^2 , initial velocity u=0m/s

Assume ‘s’ to be the height the ball is released;

(a)We know from the equation of motion;

\begin{aligned} S &=u t+1 / 2 a t^{2} \\ &=0+1 / 2(9.8)(3)^{2} \\ &=44.1 \mathrm{m} \end{aligned}

(b) Assume ‘v’ to be the velocity with which the ball strikes the ground,

We know from the equation of motion,

\begin{array}{l} v^{2}-u^{2}=2 g s \\ v^{2}-0^{2}=2(9.8)(44.1) \\ v^{2}=864.36 \\ v=29.4 \mathrm{m} / \mathrm{s} \end{array}

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