Question

If the velocity of a particle at maximum height is $\sqrt{\frac{2}{5}}$ times of it's velocity when it is at half of the greatest height. The angle of projection is

• A. ${30}^o$
• B. ${37}^o$
• C. ${60}^o$
• D. ${45}^o$

Answer 1

Answer: (C)

${60}^o$

$V_x=u\cos \theta$ (always)
$V_y^2=u_y^2-2g_y$

At max height, $u_x=u\cos \theta$
$u_y=0$

$h=\frac{u^{2}si{n}^2\theta }{2g}$
At $h^{\prime }=h/2,V_y^2=(u\sin \theta {)}^2-\frac{2gh}{2}$
$V_y^2=u^2{\sin }^2\theta -\frac{u^2{\sin }^2\theta }{2}$
$V_y=\frac{u\sin \theta }{\sqrt{2}}$

$\Rightarrow \frac{u\cos \theta }{\sqrt{(u\cos \theta {)}^2+(\frac{u\sin \theta }{\sqrt{2}}{)}^2}}=\sqrt{\frac{2}{5}}\Rightarrow \frac{{\cos }^2\theta }{{\cos }^2\theta +\frac{{\sin }^2\theta }{2}}=\frac{2}{5}$

$\Rightarrow 3{\cos }^2\theta ={\sin }^2\theta \Rightarrow \tan \theta =\sqrt{3}\Rightarrow \theta ={60}^o$