Question

A particle of mass $m$ is projected with an initial velocity $u$ at an angle $\theta$ from the ground. What is the work done by gravity during the time it reaches the highest point P?

• A. $\frac{-m{u}^2{\sin }^2\theta }{2}$
• B. $+\frac{m{u}^2{\sin }^2\theta }{2}$
• C. \N
• D. $+m{u}^2\sin \theta$

Answer 1

The correct option is A $\frac{-m{u}^2{\sin }^2\theta }{2}$

Maximum height travelled by the projectile is
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}$
Work done by gravity is dependent only on vertical displacement, hence
Work done = $\overrightarrow{F}.\overrightarrow{S}=-mg{H}_{max}=-mg\frac{u^2{\sin }^2\theta }{2g}=\frac{-m{u}^2{\sin }^2\theta }{2}$
$\therefore \left(a\right)$

Why this question? "Negative sign of work done indicates that the force opposes the motion of the particle"