For a projectile thrown into space with a speed v, the horizontal range is 3v22g. The vertical range is v28g. The angle which the projectile makes with the horizontal initially is (in degrees)

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For a projectile thrown into space with a speed v, the horizontal range is $\frac{\sqrt{3} v^{2}}{2 g} .$ The vertical range is $\frac{v^{2}}{8 g} .$ The angle which the projectile makes with the horizontal initially is (in degrees)

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answer is 30.

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Detailed Solution

$A s, \frac{v^{2} \sin 2 θ}{g} = \frac{\sqrt{3} v^{2}}{2 g}$

$o r \sin 2 θ = \frac{\sqrt{3}}{2}$

$o r 2 θ = 60^{0}$

$o r θ = 30^{0}$

Let us cross check with the help of data for vertical range.

$\frac{v^{2} \sin^{2} θ}{2 g} = \frac{v^{2}}{8 g}$

$o r \sin^{2} θ = \frac{1}{4}$

$o r \sin θ = \frac{1}{2}$

$o r θ = 30^{0}$

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