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 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..........$m$
$4.9$
$9.8$
$19.8$
$24.7$
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..........$m$
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$