A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically u

Question

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

Accepted Answer

Using the equation of motion $s = ut + \frac{1}{2}at^2$. Taking the upward direction as positive, $u = 4 	ext{ m s}^{-1}$, $a = -g = -10 	ext{ m s}^{-2}$, $t = 4 	ext{ s}$. The displacement $s = -h$ (where $h$ is the height of the bridge). So, $-h = (4)(4) + \frac{1}{2}(-10)(4)^2 = 16 - 80 = -64$. Thus, $h = 64 	ext{ m}$.

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