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

Question

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity $5 \mathrm{~m} \mathrm{~s}^{-1}$. The ball strikes the water surface after 5 s. The height of bridge above water surface is (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$)

Professor PC Thomas Classes · Answer

(a) : Given, $u=5 \mathrm{~ms}^{-1}$ (upwards), $t=5 \mathrm{~s}, g=10 \mathrm{~ms}^{-2}$ (downwards). \ Taking the downward direction as positive, initial velocity $u = -5 \mathrm{~m/s}$. \ The height of bridge above water surface is $S = ut+\frac{1}{2} g t^{2}$. \ $S = (-5)(5) + \frac{1}{2}(10)(5)^{2} = -25 + 125 = 100 \mathrm{~m}$.