A ball is dropped from a bridge of 122.5 m a | vedclass

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(C) For the first ball,the distance covered is $h = 122.5 \ m$. Using the equation of motion $h = \frac{1}{2} g t^2$:$122.5 = \frac{1}{2} 	imes 9.8 \

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$26.1$

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(C) For the first ball,the distance covered is $h = 122.5 \ m$. Using the equation of motion $h = \frac{1}{2} g t^2$:$122.5 = \frac{1}{2} 	imes 9.8 	imes t^2$$t^2 = \frac{122.5 	imes 2}{9.8} = 25$$t = 5 \ s$.The second ball is thrown $2 \ s$ later,so it has only $(5 - 2) = 3 \ s$ to reach the water.Using the equation of motion $h = ut + \frac{1}{2} g t^2$ for the second ball:$122.5 = u(3) + \frac{1}{2} 	imes 9.8 	imes (3)^2$$122.5 = 3u + 4.9 	imes 9$$122.5 = 3u + 44.1$$3u = 122.5 - 44.1 = 78.4$$u = \frac{78.4}{3} \approx 26.13 \ m/s$.

$A$ ball is dropped from a bridge of $122.5 \ m$ above a river. After the ball has been falling for $2 \ s$,a second ball is thrown straight down after it. What is the initial velocity of the second ball so that both hit the water at the same time (in $m/s$)?

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