Question

A passenger in an open car travelling at $30m{s^{ - 1}}$ throws a ball out over the bonnet. Relative to the car the initial velocity of the ball is $20m{s^{ - 1}}$ at ${60^\circ }$ to the horizontal. The angle of projection of the ball with respect to the horizontal road will be:
A) ${\tan ^{ - 1}}\left( {\dfrac{2}{3}} \right)$
B) ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)$
C) ${\tan ^{ - 1}}\left( {\dfrac{4}{{\sqrt 3 }}} \right)$
D) ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$

Answer 1

Hint: In order to find the solution of the given question, we need to find the resultant of the velocities of the car and the ball. After that we need to find the angle of projection of the velocity vectors. Then we can finally conclude with the correct solution of the given question.

Complete step by step solution:
The velocity of the car is given as,${v_c} = 30m{s^{ - 1}}$
And the velocity of ball with respect to the car is given as,${v_{bc}} = 20m{s^{ - 1}}$
Now we need to find the velocity of the ball with respect to the car with its components.
${v_{bc}} = 20\cos {60^\circ }i + 20\sin {60^\circ }j$
$ \Rightarrow {v_{bc}} = 20 \times \dfrac{1}{2}i + 20 \times \dfrac{{\sqrt 3 }}{2}j$
$\therefore {v_{bc}} = 10i + 10\sqrt 3 j$
Now, we know that,
${v_{bc}} = {v_b} - {v_c}$
$ \Rightarrow {v_b} = {v_{bc}} + {v_c}$
$ \Rightarrow {v_b} = 10i + 10\sqrt 3 j + 30i$
$\therefore {v_b} = 40i + 10\sqrt 3 j$
Now we need to find the angle of projection.
$\tan \theta = \dfrac{{10\sqrt 3 }}{{40}}$
$\therefore \theta = {\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)$
Therefore, the required value of the angle of projection is ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)$.

Hence, option (B), i.e. ${\tan ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{4}} \right)$ is the correct choice of the given question.

Note: Angle of projection: We define angle of projection as the angle from the surface with which a body is projected. For attaining the maximum range the body should be projected with an angle of ${45^\circ }$. The body can be projected both in horizontal and angular direction. When a body is thrown with an initial velocity in the horizontal direction only, then it is said to be in horizontal projection. Similarly, when the body is thrown with an initial velocity at a particular angle to the horizontal direction, then it is said to be the angular projection.
Velocity of projection: It is defined as the velocity of the body with which the body is projected.