Question

A person of height $180\,cm$ throws a ball at an angle $30^\circ$with the horizontal such that it is received by another person of similar height standing at a distance of 40 m from the first one. The maximum height attained by the ball is:
A. 6.00 m
B. 5.78 m
C. 8.00 m
D. 7.58 m

Answer 1

We have given the range of the ball. Use the expression for range of the projectile to determine the initial velocity. Once you find the initial velocity of the ball, substitute it into the expression for maximum height attained by the projectile.

Formula used:
${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Here, u is the initial velocity of the ball, $\theta$ is the angle made by the ball with the horizontal and g is the acceleration due to gravity.
The horizontal range of the projectile is,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$

Complete answer: or Complete step by step answer:
We know that the expression for a maximum height attained by a projectile is,
${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ …… (1)
Here, u is the initial velocity of the ball, $\theta$ is the angle made by the ball with the horizontal and g is the acceleration due to gravity.
Therefore, from the above equation, we need to determine the initial velocity of the ball.
We have given the maximum horizontal distance that the range of the projectile is 50 m. We know the expression for a range of projectile,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Therefore, we can substitute 40 m for R, $10\,m/{s^2}$ for g and $30^\circ$ for $\theta$ in the above equation.
$40 = \dfrac{{{u^2}\sin \left( {2 \times 30^\circ } \right)}}{{10}}$
$\Rightarrow \dfrac{{400}}{{0.866}} = {u^2}$
$\Rightarrow u = 21.49\,m/s$
Now, we can substitute 21.49 m/s for u, $10\,m/{s^2}$ for g and $30^\circ$ for $\theta$ in equation (1).
${H_{\max }} = \dfrac{{{{\left( {21.49} \right)}^2}{{\sin }^2}\left( {30^\circ } \right)}}{{2\left( {10} \right)}}$
$\Rightarrow {H_{\max }} = \dfrac{{115.46}}{{20}}$
$\Rightarrow {H_{\max }} = 5.78\,m$
This is the height attained by the ball above the height of the person. Therefore, we have to add the height of the person in the maximum height attained by the ball. Thus,
${H_{\max }} = 5.78 + 1.8$
$\therefore {H_{\max }} = 7.58\,m$.

So, the correct answer is “Option D”.

Note:
Easier way to solve this question is to take the ratio of expression for maximum height and range of the projectile. Therefore, the term initial velocity will get cancelled and you need to substitute only the angle of projection and given value of range. Note that the height of the person is given in centimeters. Therefore, convert it into meters for the calculation.