Question

A football player kicks a ball at an angle of $30^\circ$to the horizontal with an initial velocity of 15 m/s. Assuming the ball travels in a vertical plane, calculate the
(a) Maximum height
(b) Time of flight and
(c) Horizontal range ($g = 9.8\,m/{s^2}$)

Answer 1

Use the formula for maximum height, time of flight and horizontal range. Substitute the given quantities in the respective formulae to get the answer. Express the answers only in S.I units.

Formula used:

1. Maximum height, ${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$, where, u is the initial velocity, $\theta$ is the angle of projection and g is the acceleration due to gravity.
2. Time of flight, $T = \dfrac{{2u\sin \theta }}{g}$
3. Horizontal range, $R = \dfrac{{{u^2}\sin 2\theta }}{g}$

Complete step by step answer:
We have given the initial velocity of the ball, $u = 15\,m/s$ and angle of projection
$\theta = 30^\circ$.
(a) We know the formula for maximum height,
${H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Here, g is the acceleration due to gravity.
Substituting$u = 15\,m/s$, $\theta = 30^\circ$ and $g = 9.8\,m/{s^2}$ in the above equation, we get,
${H_{\max }} = \dfrac{{{{\left( {15} \right)}^2}{{\sin }^2}\left( {30^\circ } \right)}}{{2\left( {9.8} \right)}}$
$\Rightarrow {H_{\max }} = \dfrac{{56.25}}{{19.6}}$
$\therefore {H_{\max }} = 2.9\,m$

Therefore, the maximum height attained by the ball is 2.9 m.

(b) We know the formula for time of flight of a projectile motion,
$T = \dfrac{{2u\sin \theta }}{g}$
Substituting$u = 15\,m/s$, $\theta = 30^\circ$ and $g = 9.8\,m/{s^2}$ in the above equation, we get,
$T = \dfrac{{2\left( {15} \right)\sin \left( {30^\circ } \right)}}{{9.8}}$
$\Rightarrow T = \dfrac{{15}}{{9.8}}$
$\therefore T = 1.53\,s$

Therefore, the time of flight of the ball is 1.53 second.

(c) We know the formula for maximum horizontal range of the projectile,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Substituting$u = 15\,m/s$, $\theta = 30^\circ$ and $g = 9.8\,m/{s^2}$ in the above equation, we get,
$R = \dfrac{{{{\left( {15} \right)}^2}\sin 2\left( {30^\circ } \right)}}{{9.8}}$
$\Rightarrow R = \dfrac{{194.86}}{{9.8}}$
$\therefore R = 19.9\,m$

Therefore, the horizontal range of the ball is 19.9 m.

Note: Don’t get confused between the formula for maximum height and horizontal range. One can remember the formula for maximum horizontal range using the fact that the projectile attains maximum range in the horizontal direction when the angle of projection is $45^\circ$. The horizontal range will be maximum if $\sin 2\theta$ is equal to 1. Therefore, the angle should be $2\theta$ to become $90^\circ$. We know that $\sin 90^\circ = 1$. The formula for time of flight can be determined using the kinematic equation in the vertical direction of the motion.