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. $\frac{1}{2}m{u}^{2}ta{n}^2\frac{\theta }{2}$
• B. $\frac{1}{2}m{u}^{2}ta{n}^2\theta$
• C. $\frac{1}{2}m{u}^{2}co{s}^2\theta ta{n}^2\frac{\theta }{2}$
• D. $\frac{1}{2}m{u}^{2}co{s}^2\frac{\theta }{2}si{n}^2\theta$

Answer 1

The correct option is C $\frac{1}{2}m{u}^{2}co{s}^2\theta ta{n}^2\frac{\theta }{2}$


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