The horizontal range (R) of a projectile is n | vedclass

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(B) The horizontal range of a projectile is given by $R = \frac{v_0^2 \sin(2	heta_0)}{g}$.The maximum height of a projectile is given by $H = \frac{v

Vedclass · Accepted Answer

$	heta_0 = 	an^{-1}(4/n)$

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(B) The horizontal range of a projectile is given by $R = \frac{v_0^2 \sin(2	heta_0)}{g}$.The maximum height of a projectile is given by $H = \frac{v_0^2 \sin^2(	heta_0)}{2g}$.Given that $R = nH$,we substitute the formulas:$\frac{v_0^2 (2 \sin 	heta_0 \cos 	heta_0)}{g} = n \left( \frac{v_0^2 \sin^2 	heta_0}{2g} ight)$.Canceling common terms $v_0^2$,$g$,and $\sin 	heta_0$ from both sides:$2 \cos 	heta_0 = n \left( \frac{\sin 	heta_0}{2} ight)$.Rearranging to solve for $	an 	heta_0$:$\frac{\sin 	heta_0}{\cos 	heta_0} = \frac{4}{n}$.Therefore,$	an 	heta_0 = \frac{4}{n}$,which implies $	heta_0 = 	an^{-1}(\frac{4}{n})$.

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

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