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Motion in a Plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Plane3 Marks
Question
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by,
$\theta_0=\tan^{-1}\Big(\frac{4\text{H}}{\text{R}}\Big)$
where, H is the maximum height attained by the projectile and R is the range of the projectile.
✓

Answer


The path followed by a proiectile prolected at an angle $\theta_0$ with velocity $\vec{\text{u}}$ is shown in figure. The maximum height attained by the projectile is given by

$\text{H}=\frac{\text{u}^2\sin^2\theta_0}{2\text{g}}\dots(\text{i})$

The range of the projectile is given by

$\text{R}=\frac{\text{u}^2\sin2\theta_0}{\text{g}}$

$=\frac{2\text{u}^2\sin\theta_0\cos\theta_0}{\text{g}}\dots(\text{ii})$

Dividing eq. (i) by eq. (ii), we get

$\tan\theta_0=\frac{4\text{H}}{\text{R}}\Rightarrow\ \theta_0=\tan^{-1}\Big(\frac{4\text{H}}{\text{R}}\Big).$

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