Question

The coach throws a baseball to a player with an initial speed of 20 m/s at an angle of 45$^{\circ}$ with the horizontal. At the moment the ball is thrown, the player is 50 m from the coach. At what speed and in what direction must the player run to catch the ball at the same height at which it was released? (g = 10m/s$^2$)
A. $\dfrac{2}{\sqrt{3}} ms^{-1}$
B. $\dfrac{2}{\sqrt{5}} ms^{-1}$
C. $\dfrac{5}{\sqrt{2}} ms^{-1}$
D. $\dfrac{5}{\sqrt{3}} ms^{-1}$

Answer 1

As the coach throws the ball at an angle, it will undergo a projectile motion in the direction of the player. We can find the range of the ball with the help of the formula and then determine the velocity for the player as the distance required to cover is divided by the time for which the ball is in the air.

Formula used:
The range of a project tile making some angle $\theta$ with the horizontal is given as;
$R = \dfrac{u^2 \sin 2 \theta}{g}$ .
The time of flight is given as:
$T = \dfrac{2 u \sin \theta}{g}$ .

Complete answer:
We are given the initial velocity with which the ball has been projected u = 20 m/s. As the ball is launched with an angle of 45$^{\circ}$, it will cover a maximum range given by:
$R = \dfrac{u^2 \sin 2 \theta}{g} = \dfrac{u^2}{g} = \dfrac{(20)^2}{10} = 40$m.
Thus, before coming to ground the ball will travel a horizontal distance of 40 m.
The time taken for the ball to travel this distance can be written as:
$T = \dfrac{2 u \sin \theta}{g} = \dfrac{2 \times 20 \sin 45^{\circ}}{10} = 2 \sqrt{2}$ s.
Therefore this is the time for which the ball will stay in air, or this is the time that it takes for the ball to come from the coach to a horizontal distance of 40 m in the direction of the player.
But the player is present at a distance of 50 m from the coach. This means that he has to travel 10 m (50 m $-$ 40 m) distance in the time for which the ball is in the air. He must travel at a speed of:
$v = \dfrac{10}{2 \sqrt{2}} = \dfrac{5}{\sqrt{2}}$ m/s.

Therefore, the correct answer is option (C).

Note:
For any parabolic trajectory the time of ascent is equal to time of descent. Therefore we have a factor of 2 in the time of flight expression. One can also use a direct formula for the maximum horizontal range for the ball as when a projectile is projected at an angle of 45$^{\circ}$ with the horizontal we get maximum horizontal distance coverage (range) by the projectile.