A ball is thrown from a point with a speed v_0 at an angle of projection θ . From the same point and at the same instant a person starts running with a constant speed v_0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

To catch the ball, the horizontal component of the ball's velocity must be equal to the speed of the person. v_ball, x = v_person v_0 θ = v_0/2 θ = 1/2 θ = 60^ If θ = 60^ , the person remains directly under the ball throughout the flight and catches it when it returns to the same height.

A ball is thrown from a point with a speed v_0 at an angle of projection θ . From the same point and at the same instant a person starts running with - Physics MCQ AIEEE 2004 2026 | ExamTest.live

To catch the ball, the horizontal component of the ball's velocity must be equal to the speed of the person.

v_{ba ll, x} = v_{p erso n}

v_{0} cos θ = \frac{v _{0}}{2}

cos θ = \frac{1}{2} ⟹ θ = 6 0^{\circ}

θ = 6 0^{\circ}

, the person remains directly under the ball throughout the flight and catches it when it returns to the same height.

A ball is thrown from a point with a speed $v_{0}$ at an angle of projection $θ$ . From the same point and at the same instant a person starts running with a constant speed $v_{0} /2$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

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A ball is thrown from a point with a speed v0​ at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v0​/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?