Question

# One second after projection, a stone moves at an angle 45∘ with horizontal. Two seconds after projection, it moves horizontally. It's angle of projection is- [g=10 ms2]

A
tan1(3)
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B
tan1(4)
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C
tan1(3)
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D
tan1(2)
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Solution

## The correct option is D tan−1(2)Let the angle of projection be θ. Then the time taken to reach the maximum height where the velocity is horizontal, T=usinθg=2 s Hence usinθ=2g-----------(1) Now after one second the projectile makes an angle of 45∘. Hence the vertical component of velocity will be equal to horizontal component of velocity there. We know that the horizontal component of velocity remains constant in projectile which is ucosθ Hence the vertical component of velocity will also be ucosθ. Applying equation of motion after 1 sec. vy=uy+a×t ucosθ=usinθ−g×1 From equation 1 we know usinθ=2g Putting this value in the above equation ucosθ=2g−g=g--------------(2) From equation 1 and 2 tanθ=2 or θ=tan−12

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