Question

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?


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Answer 1

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By using the formula, we can find the maximum

height

$v^{2}=u^{2}+2 a s$

$0=u^2+2(-g)\times h_{\text{max }}$

$h_{\max }=\frac{u^{2}}{2 g}$

$=\frac{400}{20}$

$=20 \mathrm{~m}$

The net displacement and the total distance covered by the stone is,

$\begin{array}{l} v^{2}=u+g t \\ u=g t \\ t=\frac{20}{10} \\ =2 s \end{array}$

The maximum height reached by the stone is 20m

The net displacement is 2s.

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