Show that the projection angle _0 for a projectile launched from … - Vidyadip

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.

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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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