Question

A particle projected with an initial velocity $u$ at angle $\theta$ from the ground. What is the work done by gravity during the time it reaches the highest point $\left(P\right)$ in its path

• A. $\frac{-m{u}^2{\sin }^2\theta }{2}$
• B. $+\frac{m{u}^2{\sin }^2\theta }{2}$
• C. $0$
• D. $+m{u}^2\sin \theta$

Answer 1

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

For the work done by gravity, the gravitational force is $\left(mg\right)$.
When the particle reaches maximum height, we know that
$H_{max}=\frac{u^2{\sin }^2\theta }{2g}$
hence the work done by the gravity is
$W=-mg{H}_{\text{max}}=-mg\frac{u^2{\sin }^2\theta }{2g}=\frac{-m{u}^2{\sin }^2\theta }{2}$

(As the force of gravity and the displacement of the particle is opposite to each other we get $\theta ={180}^{\text{o}}$, where $\theta$ is the angle between force and the displacement)

Hence option A is the correct answer