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(C) Given: Initial velocity $u = 5 \,m \,s^{-1}$, Time $t = 3 \,s$, Acceleration $a = -g = -10 \,m \,s^{-2}$.Using the kinematic equation for displace

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$30$

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(C) Given: Initial velocity $u = 5 \,m \,s^{-1}$, Time $t = 3 \,s$, Acceleration $a = -g = -10 \,m \,s^{-2}$.Using the kinematic equation for displacement: $s = ut + \frac{1}{2}at^2$.Here, $s$ represents the vertical displacement from the point of projection to the water surface. Since the stone strikes the water below the bridge, the displacement is $-h$ (where $h$ is the height of the bridge).Substituting the values: $-h = (5)(3) + \frac{1}{2}(-10)(3)^2$.$-h = 15 - 5(9)$.$-h = 15 - 45$.$-h = -30$.$h = 30 \,m$.Therefore, the height of the bridge above the water surface is $30 \,m$.

If a stone thrown vertically upwards from a bridge with an initial velocity of $5 \,m \,s^{-1}$ strikes the water below the bridge in a time of $3 \,s$, then the height of the bridge above the water surface is (Acceleration due to gravity $= 10 \,m \,s^{-2}$) (in $\,m$)

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