Question
Prove that the maximum horizontal range of a projectile is four times the maximum height
✓
Answer
We know that,
Horizontal range, $\quad R =\frac{u^2 \sin 2 \theta}{g}......(1)$
For maximum $R, \quad \theta=45^{\circ}$ or $\frac{\pi}{4}$
$\begin{array}{l}
R_{\max }=\frac{u^2 \sin 2 \times \frac{\pi}{4}}{g}=\frac{u^2 \sin \frac{\pi}{2}}{g} \\
R_{\max }=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g}....(2)
\end{array}$
Maximum height, $H =\frac{u^2 \sin ^2 \theta}{2 g}$
But $\theta=45^{\circ}$
$\begin{aligned}
H & =\frac{u^2 \sin ^2 45^{\circ}}{2 g} \\
H & =\frac{u^2}{2 g} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{u^2}{2 g} \\
H & =\frac{1}{4} R_{\max } \\
\therefore \quad R_{\max } & =4 H \quad \text { Hence Proved. }
\end{aligned}$
Horizontal range, $\quad R =\frac{u^2 \sin 2 \theta}{g}......(1)$
For maximum $R, \quad \theta=45^{\circ}$ or $\frac{\pi}{4}$
$\begin{array}{l}
R_{\max }=\frac{u^2 \sin 2 \times \frac{\pi}{4}}{g}=\frac{u^2 \sin \frac{\pi}{2}}{g} \\
R_{\max }=\frac{u^2 \sin 90^{\circ}}{g}=\frac{u^2}{g}....(2)
\end{array}$
Maximum height, $H =\frac{u^2 \sin ^2 \theta}{2 g}$
But $\theta=45^{\circ}$
$\begin{aligned}
H & =\frac{u^2 \sin ^2 45^{\circ}}{2 g} \\
H & =\frac{u^2}{2 g} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{u^2}{2 g} \\
H & =\frac{1}{4} R_{\max } \\
\therefore \quad R_{\max } & =4 H \quad \text { Hence Proved. }
\end{aligned}$
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