Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

68 m

56 m

60 m

64 m

Step-by-Step Solution

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 ( $u$ ) = $+4 \text{ m s}^{-1}$ (upwards). Acceleration ( $a$ ) = $-g = -10 \text{ m s}^{-2}$ (acting downwards). Time ( $t$ ) = $4 \text{ s}$ . Displacement ( $s$ ) = $-h$ (since the water surface is below the bridge).
Apply Kinematic Equation: Use the second equation of motion : $s = ut + \frac{1}{2}at^2$
Substitute Values: $-h = (4)(4) + \frac{1}{2}(-10)(4)^2$ $-h = 16 - 5(16)$ $-h = 16 - 80$ $-h = -64$
Solve for Height: $h = 64 \text{ m}$

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