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Question: A projectile A is thrown at an angle of \(30^\circ \) to the horizontal from point P. At the same ti...

A projectile A is thrown at an angle of 3030^\circ to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2v_2 upwards from the point Q vertically below the highest point. For B to collide with A, v2v1\dfrac{{v_2}}{{v_1}} should be

A. 11
B. 22
C. 12\dfrac{1}{2}
D. 44

Explanation

Solution

The projectile A is thrown at an angle hence we will get two compnet after rev;oving. Now as per the problem statement for A and B to collide the vertical velocity of the two particles initial vertical component must be equals so as to meet at a point at a particular time. After solving two step we can get the required ratio.

Complete step by step answer:
As per the given problem a projectile A is thrown at an angle of 3030^\circ to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2v_2 upwards from the point Q vertically below the highest point.We need to calculate the v2v1\dfrac{{v_2}}{{v_1}} if the B collides with A.

As the A is projected at an angle of 3030^\circ then the velocity must be slipped into two compnet one is vertical and another is horizontal. v1v_1 is the velocity of A.
v1H=v1cos30v_1H = v_1\cos 30^\circ
Where, v1Hv_1H is the horizontal component of A.
v1V=v1sin30v_1V = v_1\sin 30^\circ
Where, v1Vv_1V is the vertical component of B.
And B is thrown with velocity v2v_2 upwards.Hence vertical velocity is the same as the projected velocity.

For A and B to collide , the initial vertical components of the velocities must be equal.
Equation velocities along the vertical we will get,
Vertical velocity of A= Vertical velocity of B
v1V=v2v_1V = v_2
Now putting the vertical component of A we will get,
v1sin30=v2v_1\sin 30^\circ = v_2
We know the value of sin30=12sin30 = \dfrac{1}{2}
Now,
v1×12=v2v_1 \times \dfrac{1}{2} = v_2
Now rearranging the equation we will get,
v2v1=12\therefore \dfrac{{v_2}}{{v_1}} = \dfrac{1}{2}

Therefore the correct option is (C)\left( C \right).

Note: When two or more bodies collide then an event called collision takes place where these bodies exert force on each other in a short period of time. Remember that the horizontal component remains the same throughout the flight and hence the horizontal motion of a projectile is independent of its vertical motion.