Question #4fa55

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Question #4fa55

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Answer 1 · 2018-01-19T14:57:19

See below

Explanation:

Using the second equation of motion,
$s = u t + \frac{1}{2} a t^{2}$ $s = ut + 1/2 at^2$
Put s = h
u = 0
$a = 10 m s^{- 2}$ $a = 10ms^-2$
We get,
$h = (0) t + \frac{1}{2} \cdot 10 \cdot t^{2}$ $h = (0)t + 1/2*10*t^2$
$h = 0 + 5 t^{2}$ $h = 0 + 5t^2$
$h = 5 t^{2}$ $h = 5t^2$ ----(1)
Thus we get a relation between distance covered by the body and time taken.
Now put $t = \frac{t}{2}$ $t = t/2$ in the equation (1)
$h' = 5(t/2)^2$
$h' = (5t^2)/4$

$h':h = (5t^2)/4 : 5t^2$
$h':h = 1:4$

Thus, the distance covered in $1/2 t$ is equal to $1/4$ h.
So, the distance of the body from where it was released is $1/4$ h.

Therefore,height of the body from ground is $3/4$ h.

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Question #4fa55

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