Question

A tennis ball is thrown up and reaches a certain height and comes down in $8s$. If value of acceleration due to gravity $g = 10m{s^{ - 2}}$, then the height reached by tennis ball and velocity with which it strikes the ground respectively is ___________ and ____________.

Answer 1

Hint:we will first calculate the time taken by the ball to reach a certain height and downward. Then we will calculate the final velocity of the tennis ball by using the suitable formula. Here, the initial velocity of the tennis ball is zero. We will calculate the height attained by the tennis ball is given below.

Formula used:
The formula used for calculating the final velocity of the tennis ball is shown below
$v = u + gt$
Here, $v$ is the final velocity of the tennis ball, $u$ is the initial velocity of the tennis ball, $g$ is the acceleration due to gravity and $t$ is the time taken.
${v^2} - {u^2} = 2gh$
Here, $h$ is the height attained by the tennis ball.

Complete step by step answer:
It is given in the question that a tennis ball is thrown up and reaches a certain height and comes down. The time taken by the tennis ball for this process is $8s$. Now, let the time taken by the tennis ball to reach a certain height is ${t_a}$ and the time taken by the tennis ball to come down is ${t_b}$. Therefore, from the condition given in the question, we get
${t_a} + {t_b} = 8$
Also, the time taken by the tennis ball to reach upward and come downward will be equal, therefore, we get
${t_a} = {t_b}$
Now, putting this relation in the above equation, we get
${t_a} + {t_a} = 8$
$ \Rightarrow \,2{t_a} = 8$
$ \Rightarrow \,{t_a} = 4$
Therefore, from the above relation, we get
${t_b} = 4$
Now, the initial velocity of the tennis ball is zero, therefore, we will calculate the final velocity as shown below
$v = u + gt$
$ \Rightarrow \,v = 0 + 10 \times 4$
$ \therefore \,v = 40\,m{s^{ - 1}}$
Now, the height attained by the tennis ball can be calculated as shown below
${v^2} - {u^2} = 2gh$
$ \Rightarrow \,h = \dfrac{{{v^2} - {u^2}}}{{2g}}$
$ \Rightarrow \,h = \dfrac{{{{\left( {40} \right)}^2} - {{\left( 0 \right)}^2}}}{{2 \times 10}}$
$ \Rightarrow \,h = \dfrac{{1600}}{{20}}$
$ \therefore \,h = 80\,m$

Therefore, the height reached by the tennis ball and velocity with which it strikes the ground respectively is $80\,m$ and $40\,m{s^{ - 1}}$.

Note: In the formula of final velocity, we have used the acceleration due to gravity instead of acceleration. Because the gravity will enable the come downward. Also, you must always remember the formulas used in the solution, you must not get confused with the formulas.