Question

A football is kicked by a player with a speed of $20\,m{s^{ - 1}}$ at an angle of projection $45^\circ$. A receiver on the goal line 25 m away from the player in the direction of the kick runs the same instant to meet the ball. What must be his speed if he is to catch the ball before it hits the ground? Ignore the height of the receiver.
(A) 2 m/s
(B) 3 m/s
(C) 4 m/s
(D) 5.3 m/s

Answer 1

Determine the horizontal range of the ball. Since the player on the goal line is at 25 m distance from the player. Therefore, find the distance between the player on the goal line and the position of the ball where it will hit the ground. Use the formula for time of flight to determine the time taken by the player on the goal line to reach the ball where it hits the ground. Use the relation between velocity, distance and time to calculate the speed of the player on the goal line.

Formula used:

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

Complete step by step answer:
First, we have to calculate the horizontal range of the ball where it hits the ground. We have the formula for horizontal range,
$R = \dfrac{{{u^2}\sin 2\theta }}{g}$
Here, u is the initial velocity of the ball, $\theta$ is the angle of projection and g is the acceleration due to gravity.
Substituting $20\,m{s^{ - 1}}$ for u, $45^\circ$ for $\theta$ and $10\,m/{s^2}$ for g in the above equation, we get,
$R = \dfrac{{{{\left( {20} \right)}^2}\sin 2\left( {45^\circ } \right)}}{{10}}$
$\Rightarrow R = \dfrac{{400}}{{10}}$
$\Rightarrow R = 40\,m$
Since the other player on the goal line is $25 m$ away from the first player, he has to move in the opposite direction of the first player to catch the ball.
The distance he travels to catch the ball is can be calculated as follows,
$d = 40\,m - 25\,m$
$\Rightarrow d = 15\,m$
We have given that the player on the goal line has started running at the instant the ball is kicked by the player. Therefore, the time taken by the player on the goal line to catch the ball can be calculated using the formula for time of flight of the projectile.
$T = \dfrac{{2u\sin \theta }}{g}$
Substituting $20\,m{s^{ - 1}}$ for u, $45^\circ$ for $\theta$ and $10\,m/{s^2}$ for g in the above equation, we get,
$T = \dfrac{{2\left( {20} \right)\sin \left( {45^\circ } \right)}}{{10}}$
$\Rightarrow T = 2.83\,s$
We can calculate the speed of the player on the goal line to catch the ball using the formula below,
$v = \dfrac{d}{t}$
Substituting 15 m for d and $2.83 s$ for t in the above equation, we get,
$v = \dfrac{{15}}{{2.83}}$
$\therefore v = 5.3\,m/s$

So, the correct answer is option (D).

Note: The value of acceleration due to gravity is $9.8\,m/{s^2}$ but for the simplicity in the calculation, we have taken it as $10\,m/{s^2}$. In the above solution, the range we calculated may vary due to air resistance force which will oppose the motion of the ball. We did not take into account the air resistance force.