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Question

A projectile is thrown with an initial velocity of (x^i+y^j) m/s. If the range of the projectile is double the maximum height reached by it, then

A
y=2x
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B
y=3x
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C
y=x
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D
y=4x
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Solution

The correct option is A y=2x
We know that, the maximum height formula is,
Hmax=u2sin2θ2g ...(ii), and the range of the projectile is given by,
R=u2sin2θg ...(ii)
It is also given that the range of the projectile is double the maximum height reached

R=2H
u2sin2θg=2×u2sin2θ2g, by simplifying we get,

tanθ=2
vyvx=2=yxy=2x

Hence option A is the correct answer

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