Question

A ball is thrown from a point with a speed ${V_0}$ at an angle of projected $\theta$. From the same point and at the same instant a person starts running with a constant speed $\dfrac{{{V_0}}}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?
$\left( A \right)Yes,60^\circ$
$\left( B \right)Yes,30^\circ$
$\left( C \right)Yes,45^\circ$
$\left( D \right)No$

Answer 1

First we need to draw a rough diagram representing all the required parameters to solve the problem. Find the time period of the projection using the time of flight formula. Now find the distance covered by the man then equate the range of projection. From that on solving we can find the angle of projection.

Complete step by step answer:
As per the problem we have, a ball is thrown from a point with a speed ${V_0}$ at an angle of projected $\theta$. From the same point and at the same instant a person starts running with a constant speed $\dfrac{{{V_0}}}{2}$ to catch the ball.
We need to find the angle of projection.
As per the first bit we need to say that the person will be able to catch the ball and the answer is yes.
Now we need to calculate the distance covered by both of them, the man as well as the particle.
For the person to calculate the ball they should travel the same distance along the horizontal axis is time T.
Time of flight of projection should be equal to the time taken by the person.
$T = \dfrac{{2{V_0}\sin \theta }}{g} \ldots \ldots \left( 1 \right)$
Where,
T is the time of flight.
$\theta$ is the angle of inclination.
${V_0}$ speed at which the ball is projected.
$g$ is the acceleration due to gravity.
Now the distance covered by the person is equal to the product of speed and time taken.
Hence,
$D = V \times T$
Where,
V is the velocity of the person.
T is the time taken by the person to cover that distance.
D is the cover by the [person that is equal to the range of the ball projected.
Now putting the respective values we will get,
$R = \dfrac{{{V_0}}}{2} \times T$
Now putting equation $\left( 1 \right)$ in above equation we will get,
$R = \dfrac{{{V_0}}}{2} \times \dfrac{{2{V_0}\sin \theta }}{g}$
Cancelling the common terms we will get,
$R = \dfrac{{{V_0}^2\sin \theta }}{g} \ldots \ldots \left( 2 \right)$
Now calculating the range of the ball or distance coned by the ball is equals to,
$R = \dfrac{{{V_0}^2\sin 2\theta }}{g} \ldots \ldots \left( 3 \right)$
Now equating equation $\left( 2 \right)$ and $\left( 3 \right)$ we will get,
$\dfrac{{{V_0}^2\sin \theta }}{g} = \dfrac{{{V_0}^2\sin 2\theta }}{g}$
Cancelling the common terms we will get,
$\sin \theta = \sin 2\theta$
We know from the trigonometric formula $\sin 2\theta = 2\sin \theta \cos \theta$
Now we will get,
$\sin \theta = 2\sin \theta \cos \theta$
Cancelling the common term we will get,
$1 = 2\cos \theta$
Rearranging the above equation we will get,
$\cos \theta = \dfrac{1}{2}$
Hence the angle of projection will be $\theta = 60^\circ$.
Therefore the correct option is (A).

Note: Remember that for a projectile motion when a ball is launched with a velocity at an angle to the horizontal of the horizontal distance travelled is called the range or the distance covered by the ball. Always you need to know that time of flight is the time from when the object is projected to the time it will take to reach the surface.