A ball is projected from a point with a speed | vedclass

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Question

(A) Let the ball be projected with speed $V_0$ at an angle $	heta$ with the horizontal. The horizontal component of the velocity of the ball is $V_x

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Let the ball be projected with speed $V_0$ at an angle $	heta$ with the horizontal. The horizontal component of the velocity of the ball is $V_x = V_0 \cos 	heta$.The person runs with a constant speed $V_p = 0.5 V_0$ in the same horizontal direction to catch the ball.For the person to catch the ball,the horizontal distance covered by the person must be equal to the horizontal range of the ball at the time of flight $T$.The horizontal range of the ball is $R = (V_0 \cos 	heta) T$,where $T = \frac{2 V_0 \sin 	heta}{g}$.The distance covered by the person in time $T$ is $d = V_p T = (0.5 V_0) T$.Equating the distance covered by the person to the horizontal range of the ball: $0.5 V_0 T = V_0 \cos 	heta T$.Dividing both sides by $V_0 T$ (assuming $T 
eq 0$),we get: $0.5 = \cos 	heta$.Therefore,$	heta = \cos^{-1}(0.5) = 60^{\circ}$.

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