A stone is dropped into water from a bridge 4 | vedclass

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Question

(A) Let $t$ be the time taken by the first stone to reach the water surface. Using the equation of motion $h = ut + \frac{1}{2}gt^2$,where $u = 0$ (dr

Vedclass · Accepted Answer

$12.25$

Vedclass · Answer

(A) Let $t$ be the time taken by the first stone to reach the water surface. Using the equation of motion $h = ut + \frac{1}{2}gt^2$,where $u = 0$ (dropped),$h = 44.1 \ m$,and $g = 9.8 \ m/s^2$:$44.1 = 0 + \frac{1}{2} 	imes 9.8 	imes t^2$$t^2 = \frac{44.1 	imes 2}{9.8} = 9$$t = 3 \ s$Since the second stone is thrown $1 \ s$ later and reaches the water at the same time,the time taken by the second stone is $t' = 3 - 1 = 2 \ s$.Using the equation $h = u't' + \frac{1}{2}g(t')^2$ for the second stone:$44.1 = u' 	imes 2 + \frac{1}{2} 	imes 9.8 	imes (2)^2$$44.1 = 2u' + 19.6$$2u' = 44.1 - 19.6 = 24.5$$u' = 12.25 \ m/s$.

$A$ stone is dropped into water from a bridge $44.1 \ m$ above the water. Another stone is thrown vertically downward $1 \ s$ later. Both strike the water simultaneously. What was the initial speed of the second stone in $m/s$?

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