Questions

# A ball is projected from the ground at angle $\mathrm{\theta }$ 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
103 ms−1
b
203 ms−1
c
105 ms−1
d
202 ms−1
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detailed solution

Correct option is

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

Similar Questions

Assertion : When the velocity of projection of a body is made n times, its time of flight becomes n times.

Reason : Range of projectile does not depend on the initial velocity of a body.

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