A ball is dropped from a bridge 122.5, metres | vedclass

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Question

Distance between the bridge and river $, S=122.5 \mathrm{m}$
The time taken by the first ball to reach the river can be determined by $:$
$s=u t+\fr

Vedclass · Accepted Answer

$26.1$

Vedclass · Answer

Distance between the bridge and river $, S=122.5 \mathrm{m}$
The time taken by the first ball to reach the river can be determined by $:$
$s=u t+\frac{1}{2} g t^{2}$
$122.5=0+\frac{1}{2} 	imes 9.8 	imes t^{2}$
$\frac{245}{9.8}=t^{2}$
$t=5 \mathrm{sec}$
As both ball reaches the water at the same time,
We can say that time taken by the second ball to reach the water is $(5-2)$ sec $=3$ sec as the second ball is thrown after $2\, sec$ of falling of first ball. The initial velocity of second ball is $:$
$s=u t+\frac{1}{2} g t^{2}$
$s=122.5 \mathrm{m}, \mathrm{t}=3 \mathrm{sec}$
$122.5=u 	imes 3+\frac{1}{2} 	imes 9.8 	imes 3^{2}$
$122.5=3 u+44.1$
$u=26.1 \mathrm{m} / \mathrm{s}$
Hence the initial velocity of the second ball so that both both ball reaches the water at the same time is $26.1 \mathrm{m} / \mathrm{s}$

A ball is dropped from a bridge $122.5\, metres$ above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. What must its initial velocity be so that both hit the water at the same time?.......$m/s$

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