A body is projected at such an angle that the horizontal range is three times the greatest height. The angle o

Question

A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is:

Accepted Answer

1.  **Formulas:** 
    *   Maximum height ($H$) is given by $H = \frac{u^2 \sin^2 	heta}{2g}$ .
    *   Horizontal range ($R$) is given by $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$ .
2.  **Given Condition:** $R = 3H$.
3.  **Substitution:**
    $\frac{2u^2 \sin 	heta \cos 	heta}{g} = 3 	imes \frac{u^2 \sin^2 	heta}{2g}$
4.  **Simplification:**
    Cancel $u^2$, $g$, and one $\sin 	heta$ (since $	heta 
eq 0$):
    $2 \cos 	heta = \frac{3}{2} \sin 	heta$
    $\frac{\sin 	heta}{\cos 	heta} = \frac{4}{3}$
    $	an 	heta = \frac{4}{3} = 1.333...$
5.  **Calculate Angle:**
    $	heta = 	an^{-1}(1.333) \approx 53.13^{\circ}$.
    Converting decimals to minutes: $0.13^{\circ} \approx 8'$.
    Therefore, $	heta = 53^{\circ}8'$.

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