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?

Question

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? (A) Yes, 60° (B) Yes,30° (C) No (D) Yes,45°

Tardigrade Admin · Accepted Answer

Yes, man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man i.e., v0 cos θ=(v0/2) where θ is the angle of projection or cos θ=(1/2) or θ=60°

Yes, $6 0^{\circ}$

Yes, $3 0^{\circ}$

No

Yes, $4 5^{\circ}$

Yes, man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man
i.e., $v_{0} cos θ = \frac{v _{0}}{2}$
where $θ$ is the angle of projection
or $cos θ = \frac{1}{2}$
or $θ = 6 0^{\circ}$

Q. 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 2v0​​ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

Yes, 60∘

Yes,30∘

No

Yes,45∘

Yes, man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man i.e., v0​cosθ=2v0​​ where θ is the angle of projection or cosθ=21​ or θ=60∘

Yes, $6 0^{\circ}$

Yes, $3 0^{\circ}$

Yes, $4 5^{\circ}$

Yes, man will catch the ball, if the horizontal component of velocity becomes equal to the constant speed of man
i.e., $v_{0} cos θ = \frac{v _{0}}{2}$
where $θ$ is the angle of projection
or $cos θ = \frac{1}{2}$
or $θ = 6 0^{\circ}$