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A stone is thrown with an initial speed of $4.9 \mathrm{~m} / \mathrm{s}$ from a bridge in vertically upward direction. It falls down in water after 2 sec . The height of bridge is
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The correct answer is:
9.8 m
Here: Initial velocity $u=-4.9 \mathrm{~m} / \mathrm{s}$ (- ve is due to vertically upward motion)
Total time $t=2 \mathrm{sec}$
The height of the bridge is given by
$s=u t+\frac{1}{2} g t^2$
$\begin{aligned} & =-4.9 \times 2+\frac{1}{2} \times 9.8 \times(2)^2 \\ & =-9.8+19.6=9.8 \mathrm{~m}\end{aligned}$
Total time $t=2 \mathrm{sec}$
The height of the bridge is given by
$s=u t+\frac{1}{2} g t^2$
$\begin{aligned} & =-4.9 \times 2+\frac{1}{2} \times 9.8 \times(2)^2 \\ & =-9.8+19.6=9.8 \mathrm{~m}\end{aligned}$
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