Question

A ball is dropped from a bridge of $122 \cdot 5{\text{m}}$ above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Find the initial velocity of the second ball such that both hit the water at the same time.
A) $49{\text{m}}{{\text{s}}^{ - 1}}$
B) ${\text{55}} \cdot {\text{5m}}{{\text{s}}^{ - 1}}$
C) ${\text{26}} \cdot {\text{1m}}{{\text{s}}^{ - 1}}$
D) $9 \cdot 8{\text{m}}{{\text{s}}^{ - 1}}$

Answer 1

Hint:The two balls are dropped from the same height. This height corresponds to the displacement of each ball and the acceleration of the two balls will be the acceleration due to gravity. As the second ball is dropped after the first ball, the second ball will take less time to reach the river. Newton’s first equation of motion can be used to find the initial velocity of the ball.

Formula used:
The displacement of a body is given by, $s = ut + \dfrac{1}{2}a{t^2}$ where $u$ is the initial velocity of the body and $a$ is the acceleration of the body.

Complete step by step answer.
Step 1: List the parameters mentioned in the question.
The displacement of the two balls is given to be $s = 122 \cdot 5{\text{m}}$ .
Let $t{\text{ sec}}$ be the time taken for the first ball to reach the river.
Then the time taken by the second ball will be $\left( {t - 2} \right){\text{sec}}$ .
The acceleration of both the balls is $a = g = 9 \cdot 8{\text{m}}{{\text{s}}^{ - 1}}$ .
Let $u$ be the initial velocity of the second ball which is to be determined.
Step 2: Using Newton’s first equation of motion we can find the time taken for the first ball to reach the river.
Newton’s first equation of motion for the first ball is given by, $s = \dfrac{1}{2}g{t^2}$ .
$ \Rightarrow {t^2} = \dfrac{{2s}}{g}$ -------- (1)
Substituting for $s = 122 \cdot 5{\text{m}}$ and $g = 9 \cdot 8{\text{m}}{{\text{s}}^{ - 1}}$ in equation (1) we get, ${t^2} = \dfrac{{2 \times 122 \cdot 5}}{{9 \cdot 8}} = 25$
Then we have the time as $t = \sqrt {25} = 5{\text{s}}$
So the first ball takes $t = 5{\text{s}}$ to reach the river.
Step 3: Using Newton’s first equation of motion we can now find the initial velocity of the second ball.
Newton’s first equation of motion for the second ball is given by,
$s = u\left( {t - 2} \right) + \dfrac{1}{2}g{\left( {t - 2} \right)^2}$ ---------- (2).
Substituting for $t = 5{\text{s}}$ , $g = 9 \cdot 8{\text{m}}{{\text{s}}^{ - 1}}$ and $s = 122 \cdot 5{\text{m}}$ in equation (2) we get, $122 \cdot 5 = u\left( {5 - 2} \right) + \dfrac{1}{2} \times 9 \cdot 8 \times {\left( {5 - 2} \right)^2}$
The above equation simplifies to $122 \cdot 5 = 3u + 44 \cdot 1$ .
$ \Rightarrow u = \dfrac{{122 \cdot 5 - 44 \cdot 1}}{3} = 26 \cdot 1{\text{m}}{{\text{s}}^{ - 1}}$
Thus the initial velocity of the second ball is $u = 26 \cdot 1{\text{m}}{{\text{s}}^{ - 1}}$ .

So the correct option is C.

Note: The first ball is considered to be under free fall and so it does not have an initial velocity i.e., $u = 0$ for the first ball and its equation of motion is obtained as $s = \dfrac{1}{2}g{t^2}$ . Though, the acceleration due to gravity is usually taken to be negative as it is directed downwards, here we neglect the sign while writing the equation of motion. This is because the initial velocity and displacement are negative too. So both sides of the equation of motion of both balls would have a negative sign if we were taking the sign into consideration.