A stone is thrown with an initial speed of 4.9 ,m/s from a bridge in vertically upward direction. It

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2.Motion in Straight Line

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A stone is thrown with an initial speed of $4.9 \,m/s$ from a bridge in vertically upward direction. It falls down in water after $2 \,sec$. The height of the bridge is..........$m$

A

$4.9$

B

$9.8$

C

$19.8$

D

$24.7$

Solution

(b) Speed of stone in a vertically upward direction is $4.9\, m/s$.

So for vertical downward motion we will consider $u = – 4.9\;m/s$

$h = ut + \frac{1}{2}g{t^2} = – 4.9 \times 2 + \frac{1}{2} \times 9.8 \times {(2)^2}$$ = 9.8\;m$

Standard 11

Physics

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