Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSMedium

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

A

4.9 m

B

9.8 m

C

19.8 m

D

24.7 m

Step-by-Step Solution

Sign Convention: Let the upward direction be positive ( $+$ ) and the downward direction be negative ( $-$ ).

Initial velocity ( $u$ ) = $+4.9 \text{ m/s}$ (upwards).
Acceleration ( $a$ ) = $-g = -9.8 \text{ m/s}^2$ (acting downwards).
Time taken ( $t$ ) = $2 \text{ s}$ .
Displacement ( $s$ ) = $-h$ (where $h$ is the height of the bridge, since the final position is below the starting point).

Apply Kinematic Equation: Use the equation $s = ut + \frac{1}{2}at^2$ . $-h = 4.9(2) + \frac{1}{2}(-9.8)(2)^2$ $-h = 9.8 + (-4.9)(4)$ $-h = 9.8 - 19.6$ $-h = -9.8$
Calculate Height: $h = 9.8 \text{ m}$

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Solved: PHYSICS Question for NEET | Sushrut