Question

A particle (mass $m$ ) is projected with an initial velocity $v$ at an angle $\theta$ to the horizontal. What is the work done by the force of gravity from the instant of throwing to the instant it reaches the highest point?

• A. $-\frac{m{v}^2{\sin }^2\theta }{2}$
• B. Zero
• C. $m{v}^2\sin \theta$
• D. $-m{v}^2\sin \theta$

Answer 1

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

Given, mass of the particle $=m$
Initial speed $=v$
In projectile motion the speed of the particle at maximum height will be only along the horizontal direction.
Final speed $=v\cos \theta$

Using work-energy theorem,
$W_{mg}=K_f-K_i=\frac{1}{2}m(v\cos \theta {)}^2-\frac{1}{2}m{v}^2$

$=\frac{1}{2}m{v}^2[{\cos }^2\theta -1]=-\frac{1}{2}m{v}^2{\sin }^2\theta$

Final answer: (d)