For a projectile thrown with a speed v- the horizontal range is -3-v-22g- The vertical range is v-28g- The angle which the projectile makes with the horizontal initially is

We know that R=v^2 sin⁡2θg According to question, v^2 sin⁡2θg= √(3)v^22g or nbsp; nbsp;sin 2θ= √(3)2 or nbsp; nbsp;2θ= 60^∘ or nbsp; nbsp;θ= ...

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Byju's Answer

Standard XII

Physics

Projectile Time, Height and Range

For a project...

Question

For a projectile thrown 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

15^{\circ}

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30^{\circ}

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45^{\circ}

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60^{\circ}

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Solution

The correct option is B $30^{\circ}$
We know that $R = \frac{v^{2} sin 2 θ}{g}$
According to question
$\frac{v^{2} sin 2 θ}{g} = \frac{\sqrt{3} v^{2}}{2 g}$
or $sin 2 θ = \frac{\sqrt{3}}{2}$ or $2 θ = 60^{\circ}$ or $θ = 30^{\circ}$
Let us cross – check with the help of data for vertical range.
$Vertical range = \frac{v^{2} {sin}^{2} θ}{2 g}$
$\frac{v^{2} {sin}^{2} θ}{2 g} = \frac{v^{2}}{8 g}$ or ${sin}^{2} θ = \frac{1}{4}$
or $sin θ = \frac{1}{2}$ or $θ = 30^{\circ}$
Hence, the correct answer is option (b)

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