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Question #189b5

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Feb 22, 2017

Let downward initial velocity of the pebble be $u$ $u$

So utilising $h = u \times t + \frac{1}{2} \times g \times t^{2}$ $h=uxxt+1/2xxgxxt^2$ we get

$91.2 = u \times 4 + \frac{1}{2} \times 9.8 \times 4^{2}$ $91.2=uxx4+1/2xx9.8xx4^2$

$\Rightarrow 4 u = 91.2 - 8 \times 9.8$ $=>4u=91.2-8xx9.8$

$\Rightarrow u = \frac{91.2}{4} - 8 \times \frac{9.8}{4}$ $=>u=91.2/4-8xx9.8/4$

$\Rightarrow u = 22.8 - 19.6 = 3.2 m/s$ $=>u=22.8-19.6=3.2"m/s"$

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