Question

A body is thrown at an angle ${\theta }_0$ with the horizontal such that it attains a speed equal to $\sqrt{\frac{2}{3}}$ times the speed of projection when the body is at half of its maximum height. Find the angle ${\theta }_0$.

Answer 1

Velocity of projectile $V=\sqrt{V_x^2+V_y^2}$

We know $V_x=V_0\cos \theta$

$V_y=V_0\sin \theta -\frac{1}{2}{gt}^2$

Also $V_y^2-V_{y_0}^2=-2gS⟶\left(1\right)$

we know that max height of projectile $H=\frac{V_0^2{\sin }^2\theta }{2g}$

$\Rightarrow \frac{H}{2}=\frac{V_0^2{\sin }^2\theta }{4g}$

Using In $\left(1\right)$ $\Rightarrow$ $V_2^2\left(atH/2\right)=V_0^2{\sin }^2\theta -2g.\frac{V_0^2{\sin }^2\theta }{4g}$

$=\frac{V_0^2{sin}^2\theta }{2}$

Given that at $H/2$ $V^2=\frac{2}{3}{V}_0^2$

$\Rightarrow \frac{V_0^2{\sin }^2\theta }{2}+V_0^2{\cos }^2\theta =\frac{2}{3}{V}_0^2$

$\Rightarrow {\sin }^2\theta +2{\cos }^2\theta =\frac{4}{3}\Rightarrow 1+{\cos }^2\theta =\frac{4}{3}$

or, ${\cos }^2\theta =\frac{1}{3}$ or $\cos \theta =\frac{1}{\sqrt{3}}$ or $\theta ={\cos }^{-1}\frac{1}{\sqrt{3}}$