Question

Two particles are projected simultaneously with the same speed $v$ in the same vertical plane with angles of elevation $\theta$ and $2\theta$, where $\theta <{45}^{\circ }$. At what time will their velocities be parallel ?

• A. $\frac{v}{g}\cos \left(\frac{\theta }{2}\right)\sin \left(\theta \right)$
• B. $\frac{v}{g}\cos \left(\frac{\theta }{2}\right)\sin \left(\frac{3\theta }{2}\right)$
• C. $\frac{v}{g}\cos \left(\frac{\theta }{2}\right)cosec\left(\frac{3\theta }{2}\right)$
• D. $\frac{v}{g}\cos \left(\frac{\theta }{2}\right)cosec\left(\frac{\theta }{2}\right)$

Answer 1

The correct option is C $\frac{v}{g}\cos \left(\frac{\theta }{2}\right)cosec\left(\frac{3\theta }{2}\right)$

Velocity of a particle projected with initial velocity $v$ at angle $\theta$ with the horizontal after time $t$
$\overrightarrow{v_1}=v_x\hat{i}+v_y\hat{j}=v\cos \theta \hat{i}+(v\sin \theta -gt)\hat{j}$

Velocity of particle projected at angle $2\theta$ after time $t$
$\overrightarrow{v_2}=v_x^{\prime }\hat{i}+v_y^{\prime }\hat{j}=v\cos 2\theta \hat{i}+(v\sin 2\theta -gt)\hat{j}$

Since the velocities are parallel,
$\frac{v_y}{v_x}=\frac{v_y^{\prime }}{v_x^{\prime }}\Rightarrow \frac{v_x}{v_x^{\prime }}=\frac{v_y}{v_y^{\prime }}\Rightarrow \frac{v\cos \theta }{v\cos 2\theta }=\frac{v\sin \theta -gt}{v\sin 2\theta -gt}$

Solving above equation, we get
$vg(\cos 2\theta -\cos \theta )t=v^2(\sin \theta \cos 2\theta -\cos \theta \sin 2\theta )$
$\Rightarrow t=\frac{v}{g}\times \frac{1}{-2\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}(-\sin \theta )$

$\Rightarrow t=\frac{v}{g}\times \frac{-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{-2\sin \frac{3\theta }{2}\sin \frac{\theta }{2}}$

$\Rightarrow t=\frac{v}{g}\cos \left(\frac{\theta }{2}\right)\text{cosec}\left(\frac{3\theta }{2}\right)$