Question

A ball is dropped from a diving board $4.9\,m$ above a lake. It hits the water with velocity $v$ and then sinks to the bottom with constant velocity $v$. It reaches the bottom in $5.0\,s$ after it is dropped. The average velocity of the ball is
A. $8.82\,m{s^{ - 1}}$
B. Zero
C. $7.5\,m{s^{ - 1}}$
D. $25\,m{s^{ - 1}}$

Answer 1

Hint: We will first calculate the time taken by the ball to reach the water. Then we will calculate the velocity that the ball will acquire when it is thrown on the surface. After the ball will go into the water, we will calculate the time taken by the ball to reach the bottom of the lake. Then, we will calculate the velocity of the ball after it goes into water. Also, we will calculate the depth of the lake to calculate the average velocity of the ball.

Formula used:
The formula used for calculating the time taken by the ball to reach the water is given below
$t = \sqrt {\dfrac{{2h}}{g}} $
Here, $t$ is the time period, $h$ is the height of the ball from the water and $g$ is the acceleration due to gravity.
The formula for calculating the velocity acquired by the ball to reach the water is given below
$v = gt$
Here, $v$ is the velocity of the ball, $g$ is the acceleration due to gravity and $t$ is the time taken by the ball to reach the water.

Complete step by step answer:
The time taken by the ball to reach the water can be calculated as shown below
${t_1} = \sqrt {\dfrac{{2h}}{g}} $
$ \Rightarrow \,{t_1} = \sqrt {\dfrac{{2 \times 4.9}}{{9.8}}} $
$ \Rightarrow \,{t_1} = 1.0s$
Now, the velocity acquired by the ball to reach the water is given below
$v = g{t_1}$
$ \Rightarrow \,v = 9.8 \times 1.0$
$ \Rightarrow \,v = 9.8\,m{s^{ - 1}}$
Now, the time taken by the ball to reach the bottom of the lake from the water surface is given below
${t_2} = 5.0 - 1.0 = 4.0s$
Now, it is given in the question that the velocity of the ball in the water is constant, therefore, the depth of the lake will be
$d = v{t_2}$
$ \Rightarrow \,d = 9.8 \times 4.0s$
$ \Rightarrow \,d = 39.2m$
Now, the average velocity of the ball will be
$\left\langle v \right\rangle = \dfrac{{total\,displacement}}{{total\,time}}$
$ \Rightarrow \,\left\langle v \right\rangle = \dfrac{{4.9 + 39.2}}{{5.0}}$
$ \therefore \,\left\langle v \right\rangle = 8.82\,m{s^{ - 1}}$
Therefore, the average velocity of the ball is $8.82\,m{s^{ - 1}}$.

Hence, option A is the correct option.

Note:We have calculated the depth of the lake, so that we can calculate the average velocity of the ball. Here, the displacement of the ball is the sum of the distance covered by the ball to reach the water surface and the distance covered by the ball to reach the bottom of the lake from the water surface. Average velocity is the total velocity acquired by the ball to reach the water surface and the velocity acquired by the ball to reach the bottom of the lake.