A horizontal bridge is built across a river. | vedclass

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(D) Given:Initial velocity $u = 4\,m s^{-1}$ (upwards is taken as positive).Time $t = 4\,s$.Acceleration $a = -g = -10\,m s^{-2}$ (downwards is taken

Vedclass · Accepted Answer

$64$

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(D) Given:Initial velocity $u = 4\,m s^{-1}$ (upwards is taken as positive).Time $t = 4\,s$.Acceleration $a = -g = -10\,m s^{-2}$ (downwards is taken as negative).Let the height of the bridge be $H$. The displacement $S$ of the ball when it hits the water surface is $-H$ (since it ends up below the starting point).Using the equation of motion $S = ut + \frac{1}{2}at^2$:$-H = (4)(4) + \frac{1}{2}(-10)(4)^2$$-H = 16 - 5 	imes 16$$-H = 16 - 80$$-H = -64$$H = 64\,m$Thus,the height of the bridge above the water surface is $64\,m$.

$A$ horizontal bridge is built across a river. $A$ student standing on the bridge throws a small ball vertically upwards with a velocity $4\,m s^{-1}$. The ball strikes the water surface after $4\,s$. The height of the bridge above the water surface is $......\,m$ (Take $g=10\,m s^{-2}$).

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