A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically u

Question

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity $4 	ext{ m s}^{-1}$. The ball strikes the water surface after $4 	ext{ s}$. The height of bridge above water surface is: (Take $g=10 	ext{ m s}^{-2}$)

Accepted Answer

1.  **Define Coordinate System:** Let the upward direction be positive ($+$) and the downward direction be negative ($-$). The origin is the point of projection (bridge).
    *   Initial velocity ($u$) = $+4 	ext{ m s}^{-1}$ (upwards).
    *   Acceleration ($a$) = $-g = -10 	ext{ m s}^{-2}$ (acting downwards).
    *   Time ($t$) = $4 	ext{ s}$.
    *   Displacement ($s$) = $-h$ (since the water surface is below the bridge).
2.  **Apply Kinematic Equation:** Use the second equation of motion :
    $$s = ut + \frac{1}{2}at^2$$
3.  **Substitute Values:**
    $$-h = (4)(4) + \frac{1}{2}(-10)(4)^2$$
    $$-h = 16 - 5(16)$$
    $$-h = 16 - 80$$
    $$-h = -64$$
4.  **Solve for Height:**
    $$h = 64 	ext{ m}$$

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