Question
Two bodies are thrown with the same initial velocity at angles $\alpha$ and $(90^\circ-\alpha)$ with and is of horizontal. What will be the ratio of (i) maximum heights attained by them and (ii) of horizontal ranges?
✓
Answer
Horizontal range, $\text{R}=\frac{\text{u}^2}{\text{g}}\sin2\theta$
and Max. height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
Case (i): When $\theta=\alpha;$
$\text{R}_1=\frac{\text{u}^2}{\text{g}}\sin2\alpha$
and $\text{H}_1=\frac{\text{u}^2\sin^2\alpha}{2\text{g}}$
Case (ii): When $\theta=(90^\circ-\alpha);$
$\text{R}_2=\frac{\text{u}^2\sin2(90^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2\sin2\alpha}{\text{g}}$
and $\text{H}_2=\frac{\text{u}^2\sin^2(90^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2\cos^2\alpha}{\text{g}}$
$\therefore\ \frac{\text{H}_1}{\text{H}_2}=\frac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha$ and $\frac{\text{R}_1}{\text{R}_2}=1.$
and Max. height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$
Case (i): When $\theta=\alpha;$
$\text{R}_1=\frac{\text{u}^2}{\text{g}}\sin2\alpha$
and $\text{H}_1=\frac{\text{u}^2\sin^2\alpha}{2\text{g}}$
Case (ii): When $\theta=(90^\circ-\alpha);$
$\text{R}_2=\frac{\text{u}^2\sin2(90^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2\sin2\alpha}{\text{g}}$
and $\text{H}_2=\frac{\text{u}^2\sin^2(90^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2\cos^2\alpha}{\text{g}}$
$\therefore\ \frac{\text{H}_1}{\text{H}_2}=\frac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha$ and $\frac{\text{R}_1}{\text{R}_2}=1.$
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