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

Question

Dr. Abubacker · Accepted Answer

(b) $\frac{v^{2}\sin{2	heta}}{g} = \frac{\sqrt{3}v^{2}}{2g}$
or $\sin{2	heta} = \frac{\sqrt{3}}{2}$ or $2	heta = 60{^\circ}$ or$\ 	heta = 30{^\circ}$
Let us cross-check with the help of data for vertical range
$\frac{v^{2}\sin^{2}	heta}{2g} = \frac{v^{2}}{8g}\ \ \ or\ \ \ \sin^{2}{	heta = \frac{1}{4}}$
$or\ \ sin\ 	heta = \frac{1}{2}\ \ \ or\ \ \ 	heta = 30{^\circ}$

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

Explanation

Reviewed by Dr. Abubacker

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