Question

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

• A. 60 m
• B. 64 m
• C. 68 m
• D. 56 m

Answer 1

The Correct Option is B

To determine the height of the bridge above the water surface, we must analyze the motion of the ball. The ball is thrown vertically upward with an initial velocity \(u = 4\ ms^{-1}\) and later strikes the water surface after 4 seconds.

We will use the equation of motion:

\(s = ut + \frac{1}{2} a t^2\)

Where:

• \(s\) is the displacement
• \(u=4\ ms^{-1}\) is the initial velocity
• \(a=-g=-10\ ms^{-2}\) is the acceleration due to gravity (negative because it acts downwards)
• \(t=4\ s\) is the time duration

Substituting these values into the equation:

\(s = 4 \times 4 + \frac{1}{2} \times -10 \times (4)^2\)

\(s = 16 - 80\)

\(s = -64 \ \text{m}\)

The negative sign indicates that the displacement is in the downward direction, i.e., the distance from the bridge to the water surface.

Thus, the height of the bridge above the water surface is 64 m.

Hence, the correct answer is: 64 m.