A ball is dropped from a bridge that is 45 m | vedclass

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(C) Given: Height of the bridge $h = 45 \ m$. Initial velocity of the ball $u = 0 \ m \ s^{-1}$. Acceleration due to gravity $g = 10 \ m \ s^{-2}$.Usi

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(C) Given: Height of the bridge $h = 45 \ m$. Initial velocity of the ball $u = 0 \ m \ s^{-1}$. Acceleration due to gravity $g = 10 \ m \ s^{-2}$.Using the kinematic equation for vertical motion: $h = ut + \frac{1}{2}gt^2$.Since $u = 0$, we have $45 = 0 + \frac{1}{2} 	imes 10 	imes t^2$.$45 = 5t^2 \implies t^2 = 9 \implies t = 3 \ s$.The ball takes $3 \ s$ to reach the water surface.During this time, the boat must travel the distance of $12 \ m$ to be at the point of impact.Speed of the boat $v = \frac{	ext{distance}}{	ext{time}} = \frac{12 \ m}{3 \ s} = 4 \ m \ s^{-1}$.

$A$ ball is dropped from a bridge that is $45 \ m$ above the water. It falls directly into a boat which is moving with constant velocity. The boat is $12 \ m$ away from the point of impact when the ball is dropped. The speed of the boat is (Take $g=10 \ m \ s^{-2}$) (in $m \ s^{-1}$)

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