A projectile is thrown with an initial velocity of v=ai^+bj^ , if the range of projectile is double the maximum height reached by it then

R=2Hmax υ2sin2θg=υ2sin2θg2sinθcosθ=sin2θ 2=tanθ b=2a

A projectile is thrown with an initial velocity of v=ai^+bj^ , if the range of projectile is double the maximum height reached by it then

R=2Hmax υ2sin2θg=υ2sin2θg2sinθcosθ=sin2θ 2=tanθ b=2a

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A projectile is thrown with an initial velocity of $v = a \hat{i} + b \hat{j}$ , if the range of projectile is double the maximum height reached by it then

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b = 4a

a = 2b

b = a

b = 2a

answer is C.

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Detailed Solution

$R = 2 H max$

$\frac{υ^{2} \sin 2 θ}{g} = \frac{υ^{2} \sin^{2} θ}{g}$

$2 \sin θ \cos θ = \sin^{2} θ$

$2 = \tan θ$ $b = 2 a$

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