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(C) Let the velocity of the car be $v_c = 30\,m/s$. The ball is thrown vertically upward relative to the car with velocity $u_y$. The time of flight $

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both $(A)$ and $(B)$

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(C) Let the velocity of the car be $v_c = 30\,m/s$. The ball is thrown vertically upward relative to the car with velocity $u_y$. The time of flight $T$ is the time taken by the car to cover $240\,m$,so $T = \frac{d}{v_c} = \frac{240}{30} = 8\,s$.For the ball,the time of flight is $T = \frac{2u_y}{g}$. Substituting $T = 8\,s$ and $g = 10\,m/s^2$,we get $8 = \frac{2u_y}{10}$,which gives $u_y = 40\,m/s$.Relative to the car,the velocity is $40\,m/s$ at an angle of $90^o$ (vertically upward).Relative to the ground,the velocity components are $u_x = 30\,m/s$ and $u_y = 40\,m/s$. The resultant speed is $u = \sqrt{u_x^2 + u_y^2} = \sqrt{30^2 + 40^2} = 50\,m/s$.The angle of projection with the horizontal is $	heta = 	an^{-1}(\frac{u_y}{u_x}) = 	an^{-1}(40/30) = 	an^{-1}(4/3)$.

$A$ person is standing on an open car moving with a constant velocity of $30\,m/s$ on a straight horizontal road. The man throws a ball in the vertically upward direction and it returns to the person after the car has moved $240\,m$. Find the speed and the angle of projection.

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