Question

A particle of mass 'm' is projected with velocity 'u' at an angle $\theta$ with horizontal. During the period when the particle descends from highest point to the position where its velocity vector makes an angle $\frac{\theta }{2}$ with horizontal, work done by the gravity force is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C


As horizontal component of velocity does not change $vcos\frac{\theta }{2}=ucos\theta$
$v=\frac{ucos\theta }{cos\frac{\theta }{2}}$
$(W_{gravity}=\Delta K=\frac{1}{2}m{v}^2-\frac{1}{2}m(ucos\theta {)}^2$

$\frac{1}{2}m[{\left(u\frac{\cos \theta }{\cos \frac{\theta }{2}}\right)}^2-{\left(u\cos \theta \right)}^2]$

$\frac{1}{2}m{u}^2{\cos }^2\theta (\frac{1}{{\cos }^2\frac{\theta }{2}}-1)$

$=\frac{1}{2}m{u}^2{\cos }^2\theta \left(\frac{1-{\cos }^2\frac{\theta }{2}}{{\cos }^2\frac{\theta }{2}}\right)$

$=\frac{1}{2}m{u}^{2}co{s}^2\theta ta{n}^2\frac{\theta }{2}$