Question

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

Answer 1

Hint: Here we use the second equation of motion $\Delta S = ut + \dfrac{1}{2}g{t^2}$, where $u$ is the initial velocity, $g$ is the acceleration due to gravity, $t$ is the time taken and $\Delta S$ is the change in motion or displacement.

Complete step by step answer:
Equation of motion describes a body’s location, velocity or acceleration relative to a given reference fame. Newton’s second law which states that the force acting on a body is proportional to the body’s mass $m$ multiplied by its acceleration, $F = ma$ is the fundamental equation of motion in classical mechanics. The velocity and position of the body as time functions will technically be derived from Newton’s equation by a method known as integration when the force acting on a body is known as a function of time. Velocity is the time scale of position $S$ transition and therefore the velocity equation integration results in $\Delta S = \dfrac{1}{2}g{t^2}$.
Given,
Ball dropped from a bridge of height, $\Delta S = 122.5\,m$
The ball falls from the bridge for time = $2\,s$
For the first ball let the time taken be $t$.
Here there is no initial velocity.
Also acceleration due to gravity, $g = 9.8\,m/{s^2}$
So,
$\Delta S = \dfrac{1}{2}g{t^2}$
$122.5 = \dfrac{1}{2} \times g \times {t^2}$ ...... (1)
According to question, for the second ball the time taken is $t - 2\,s$ .
$\Delta S = ut + \dfrac{1}{2}g{t^2}$
$122.5 = u(t - 2) + \dfrac{1}{2}g{(t - 2)^2}$ ...... (2)
$\dfrac{{g{t^2}}}{2} = ut - 2u + \dfrac{{g{t^2}}}{2} + 2g - 2gt$
$(2g - u)t = 2g - 2u$
$t = \dfrac{{2g - 2u}}{{2g - u}}$

Substituting the values in equation (1) we get,
$
{t^2} = \dfrac{{2 \times 122.5}}{g} \\
t = \sqrt {\dfrac{{2 \times 122.5}}{{98}}} = \dfrac{{35}}{7} = 5\,s \\
\Rightarrow 10g - 5u = 2g - 2u \\
\Rightarrow 8g = 3u \\
u = \dfrac{{8 \times 9.8}}{3} = 26.1\,m/s \\
$
Hence, option C is correct.
So, the initial velocity of the second ball so that both the balls hit the water at the same time is $26.1\,m/s$

Note:It is very easy to get confused whether to take initial velocity or not. The initial velocity will be zero if the motion starts from rest and the frame of reference is the same.