Question

A particle of mass 'm' is projected with a velocity 'u' at an angle ${}^{\prime }{\theta }^{\prime }$ with the horizontal. Work done by gravity during its descent from its highest point to a point which is at half the maximum height is?

• A. none of these
• B. $\frac{m{u}^2{\sin }^2\theta }{4}$
• C. $\frac{1}{2}m{u}^2{\sin }^2\theta$
• D. $\frac{1}{2}m{u}^2{\cos }^2\theta$

Answer 1

The correct option is A none of these

workdone = - change in potential.

initial potential energy,${P.E}_i=mgh=m\ast (-g)\ast o=0$

Final potential energy,=MgH

$H=\frac{u^2{sin}^2\theta }{2g}$

$P.E\left(final\right)=\frac{{Mgu}^2{sin}^2\theta }{2g}$

$Change=P.E\left(final\right)-P.E=\frac{{Mgu}^2{sin}^2\theta }{2g}-0=\frac{{Mgu}^2{sin}^2\theta }{2g}$