Question

A ball is released from a height of 1 m. what time it will take to reach the surface of the earth and how?
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Answer 1

$\sf \fcolorbox{red}{pink}{ \huge{Solution :)}}$

Given ,

• initial speed (u) = 0 m/s
• height (h) = 1 m
• acceleration due to gravity (g) = 9.8 m/s²

By using Newton's second equation of motion

$\large \sf \fbox{s= ut + \frac{1}{2} a {(t)}^{2}}$

Substitute the known values , we get

$\sf \mapsto 1 = 0 \times t + \frac{1}{2} (9.8) {(t)}^{2} \\ \\ \sf \mapsto 1 = 4.9 {(t)}^{2} \\ \\ \sf \mapsto {(t)}^{2} = \frac{1}{4.9} \\ \\ \sf \mapsto t = 0.45 sec$

Hence , the time at which it reaches the ground is 0.45 sec

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