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Projection Under uniform Acceleration

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?


Suppose the angle made by the instantaneous velocity with the horizontal be α.then

tanα=vyvx=usinθ-gtucosθ  Given that α=45°, when t=1 s;α=0°, when t=2 s  This gives ucosθ=usinθ-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

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