A ball is projected from the ground at angle θ with the horizontal. After $$1$$ s, it is moving at angle $$45$$° with the horizontal and after $$2$$ s it is moving horizontally. What is the velocity of projection of the ball?

Question

A ball is projected from the ground at angle θ with the horizontal. After $$1$$ s, it is moving at angle $$45$$° with the horizontal and after $$2$$ s it is moving horizontally. What is the velocity of projection of the ball?

Infinity Learn · Accepted Answer

Suppose the angle made by the instantaneous velocity with the horizontal be α.thentanα$$=vyvx=usin$$θ$$-gtucos$$θ  Given that α$$=45$$°, when $$t=1$$ s;α$$=0$$°, when $$t=2$$ s.   This gives ucosθ$$=usin$$θ$$-g$$  and   usinθ$$-2$$ $$g=0$$  Solving Eqs. (i) and (ii), we find usinθ$$=2g$$ and ucosθ$$=g$$. Squaring and adding,  $$u=5$$ $$g=105$$ $$ms-1$$

A ball is projected from the ground at angle $θ$ with the horizontal. After 1 s, it is moving at angle 45° with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball?

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A ball is projected from the ground at angle θ with the horizontal. After 1 s, it is moving at angle 45° with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball?

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A ball is projected from the ground at angle $θ$ with the horizontal. After 1 s, it is moving at angle 45° with the horizontal and after 2 s it is moving horizontally. What is the velocity of projection of the ball?