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Question

The trajectory of a particle in a vertical plane is 4y=3xx24, x and y are the horizontal and vertical distances of the projectile from the point of projection. The angle of projection from the horizontal and the maximum height attained by the particle are

A
45,34 units
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B
53,92 units
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C
37,94 units
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D
60,916units
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Solution

The correct option is C 37,94 units
Given, equation y=34xx216
Comparing the above with the equation of projectile
y=xtanθgx22u2cos2θ
tanθ=34θ=tan1(34)=37

(Angle of projection)
We know that maximum height of projectile (H)=u2sin2θ2g
From equation 2u2cos2θg=16

Multiplying and dividing the equation of maximum height

H=u2sin2θ2g×tan2θtan2θ=u2cos2θ2g×tan2θ
=(2u2cos2θg)×tan2θ4
=16×(34)24=4×916=94 units

Hence option C is the correct answer

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