A ball is dropped from a bridge 122.5, m abov | vedclass

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Question

(C) The distance between the bridge and the river is $S = 122.5\, m$.For the first ball,the time taken to reach the river is calculated using $S = ut

Vedclass · Accepted Answer

$26.1$

Vedclass · Answer

(C) The distance between the bridge and the river is $S = 122.5\, m$.For the first ball,the time taken to reach the river is calculated using $S = ut + \frac{1}{2}gt^2$:$122.5 = 0 + \frac{1}{2} 	imes 9.8 	imes t^2$$t^2 = \frac{122.5 	imes 2}{9.8} = 25$$t = 5\, s$.Since the second ball is thrown $2\, s$ after the first,it has $(5 - 2) = 3\, s$ to reach the water.For the second ball,using $S = ut + \frac{1}{2}gt^2$ with $S = 122.5\, m$ and $t = 3\, s$:$122.5 = u(3) + \frac{1}{2} 	imes 9.8 	imes (3)^2$$122.5 = 3u + 44.1$$3u = 122.5 - 44.1 = 78.4$$u = \frac{78.4}{3} \approx 26.13\, m/s$.Rounding to the nearest provided option,the initial velocity is $26.1\, m/s$.

$A$ ball is dropped from a bridge $122.5\, m$ above a river. After the ball has been falling for $2\, s$,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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