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Question

(b)

Given, velocity of projection of apple, $v=24 \,m / s$

Distance between point of projection and hoop is given by

$A C=\sqrt{(25)^2+(45)^2

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(b)

Given, velocity of projection of apple, $v=24 \,m / s$

Distance between point of projection and hoop is given by

$A C=\sqrt{(25)^2+(45)^2}=\sqrt{2650} m$

So, time taken by the ball to the hoop,

$t=\frac{\sqrt{2650}}{24}$ $s$

The distance covered by hoop is

$s=\frac{1}{2} g t^2=\frac{1}{2} 	imes 10 	imes(\frac{\sqrt{2650}}{{24}})^2$

$=5 	imes \frac{2650}{576}=\frac{13250}{576}$

$	herefore$ Height above the ground where apple go through the hoop is

$H=45-\frac{13250}{576}$

$=\frac {12670}{576}\simeq 22 \,m$

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