Q.

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?

Moderate

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103 ms−1

203 ms−1

105 ms−1

202 ms−1

answer is C.

(Detailed Solution Below)

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Detailed Solution

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