Question

A ball is dropped from a bridge of $122.5m$ above a river. After the ball has been falling for $2s$, a second ball is thrown straight down after it. What should be the initial velocity of the second ball be so that both hit the water at the same time?
$\left( A \right) 40m/s$
$\left( B \right) 55.5m/s$
$\left( C \right) 26.1m/s$
$\left( D \right) 9.6m/s$

Answer 1

Hint: Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Force is proportional to mass and acceleration. The velocity and the position can be derived from the newton equation by the method of integration. Here force acting on a body is known as the function of time.

Formula used:
$s = ut + \dfrac{1}{2}g{t^2}$
$u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity

Complete step by step solution:
Equation of motion helps to describe a body’s location, velocity, or acceleration relative to the frame of reference. The velocity and the position can be derived from the newton equation by the method of integration. Here force acting on a body is known as the function of time. The velocity equation integration results in the distance equation. Here $u = 0$,$t = 2s$,$s = 122.5m$.
$s = ut + \dfrac{1}{2}g{t^2}$
$s = \dfrac{1}{2}g{t^2}$
$122.5 = \dfrac{1}{2}g{t^2} - - - - - \left( 1 \right)$
For the second ball, the time taken is $(t - 2)s$
$122.5 = u\left( {t - 2} \right) + \dfrac{1}{2}g{\left( {t - 2} \right)^2}$
$t = \dfrac{{2g - 2u}}{{2g - u}} - - - - - \left( 2 \right)$
Substitute the values in equation (1),
${t^2} = \dfrac{{2 \times 122.5}}{g}$
$t = 5s$
Then substitute in equation (2)
$\dfrac{{2g - 2u}}{{2g - u}} = 5$
\[ \Rightarrow 2g - 2u = 5(2g - u)\]
$ \Rightarrow u(5 - 2) = 2g(5 - 1)$
$ \Rightarrow u = \dfrac{{2g \times 4}}{3}$
$ \Rightarrow u = \dfrac{{8 \times 9.8}}{3} = 26.1m/s$

Hence option $(C)$ is the right option.

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity and the position can be derived from the newton equation by the method of integration. The velocity equation integration results in the distance equation.