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A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity $4\ ms^{-1}$. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take $g=10\ ms^{-2}$):
A
68 m
B
56 m
C
60 m
D
64 m
Explanation
Taking upward as positive, $S=-h$, so $-h=ut-\frac{1}{2}gt^2$ gives $h=64\ m$.
Detailed Solution
Let height of bridge $=h$. Displacement of ball, $S=-h$
$S=ut+\frac{1}{2}at^2$
$-h=4\times4+\frac{1}{2}(-10)(4)^2$
$\Rightarrow h=64\ m$