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Question

The equations of motion of a projectile are given by x=36tm and 2y=96t−9.8t2m. The angle of projection with the horizontal is:

A
sin1(45)
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B
sin1(35)
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C
sin1(43)
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D
sin1(34)
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Solution

The correct option is A sin1(45)
Given : x=36t
and 2y=96t9.8t2
or y=48t4.9t2
Let the initial velocity of projectile be u and angle of projection is θ. Then, Initial horizontal component of velocity,
ux=ucosθ=(dxdt)t=0=36 ....(i)
orusinθ=(dxdt)t=0=48 ...(ii)
Dividing (ii) by (i), we get
tanθ=4836=43
sinθ=45 or θ=sin1(45)


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