Question
Prove the following statement, “For Elevation which exceed or fall short of 45° by equal amount, the range is equal.”
✓
Answer
At 45°, the projectile has maximum range.
At $(45^\circ-\theta)$ the range is,
$\text{R}_1=\frac{\text{u}^2\sin[2(45^\circ-\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ-2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
At $(45^\circ+\theta)$ the range is,
$\text{R}_2=\frac{\text{u}^2\sin[2(45^\circ+\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ+2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
$\text{R}_1=\text{R}_2$
At $(45^\circ-\theta)$ the range is,
$\text{R}_1=\frac{\text{u}^2\sin[2(45^\circ-\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ-2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
At $(45^\circ+\theta)$ the range is,
$\text{R}_2=\frac{\text{u}^2\sin[2(45^\circ+\theta)]}{\text{g}}$
$=\frac{\text{u}^2\sin(90^\circ+2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$
$\text{R}_1=\text{R}_2$
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