Question

What is the angle of projectile with the vertical if the velocity at the highest point is $\sqrt{2/5}$ times the velocity when it is at a height equal to half of the maximum height?

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${30}^{\circ }$

$v_y^2-v^2{\sin }^2\theta =-2g\left[\frac{1}{2}\frac{v^2{\sin }^2\theta }{2g}\right]$
or $v_y^2=v^2{\sin }^2\theta -\frac{v^2{\sin }^2\theta }{2}=\frac{v^2{\sin }^2\theta }{2}$
or $v_y=\frac{v\sin \theta }{\sqrt{2}}$
$v^{\prime 2}=(v\cos \theta {)}^2+{\left(\frac{v\sin \theta }{\sqrt{2}}\right)}^2$
$=v^2{\cos }^2\theta +\frac{v^2{\sin }^2\theta }{2}$
$=v^2({\cos }^2\theta +\frac{{\sin }^2\theta }{2})$
$v^{\prime }=v\sqrt{{\cos }^2\theta +\frac{{\sin }^2\theta }{2}}$
$v^2{\cos }^2\theta =\frac{2}{5}{v}^2[{\cos }^2\theta +\frac{{\sin }^2\theta }{2}]$
or $5{\cos }^2\theta =2{\cos }^2\theta +{\sin }^2\theta$ or $3{\cos }^2\theta ={\sin }^2\theta$
or ${\tan }^2\theta =3$ or $\tan \theta =\sqrt{3}$ or $\theta ={60}^{\circ }$
Angle of projection with vertical is ${90}^{\circ }-{60}^{\circ }={30}^{\circ }$.