A body is projected at an angle $30^{\circ}$ with the horizontal with velocity $v_1$ from the point

Question

A body is projected at an angle $30^{\circ}$ with the horizontal with velocity $v_1$ from the point $A$. At the same time, another body is projected vertically upwards from $B$ with velocity $v_2$. The point $B$ lies vertically below the highest point of the trajectory of body projected from point $A$. For both the bodies to collide, the value of $\frac{v_1}{v_2}$ should be $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(2) : The two bodies will collide at the highest point if both cover the same vertical height in the same time. So \ $ \frac{v_1^2 \sin ^2 30^{\circ}}{2 g}=\frac{v_2^2}{2 g} 	ext { or } \frac{v_2}{v_1}=\sin 30^{\circ}=\frac{1}{2} ; 	herefore \frac{v_1}{v_2}=2 $