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

We know that:
\(S=ut+\frac{1}{2}at^2\)
\(-H=4\times4-\frac{1}{2}\times10\times4^2\)
\(-H=16-80\)
\(H=64\,m\)

So, the correct option is (B): \(60 m\)

Answer 2

Given: 

• A small ball is thrown vertically upwards with a velocity of 4 m/s from a horizontal bridge.
• The ball strikes the water surface after 4 seconds.
• Acceleration due to gravity,
  \(g = 10 m/s^2\)
  \(u = 4m/s\)
  \(g = -10 m/s^2\)

Let's assume the initial height of the ball above the water surface to be h.

We know that the distance traveled by the ball in the upward direction is equal to that traveled by the ball in the downward direction.
t = 4 seconds

Displacement= -ve, and let this displacement be =S

\(S = Ut + \frac{1}{2} \times a(t^2)\)
\(-S = 4 . 4 + \frac{1}{2} \times (-10) \times16\)
\(-S = 16 - 16(5)\)
\(-S = -64\)
\(S = 64 m\)

Hence The height of the bridge above the water surface is 64 meters.