Question

A projectile is fired with some velocity making certain angle with the horizontal. Which of the following graphs is the best representation for the kinetic energy of a projectile $\left(KE\right)$ versus its horizontal displacement $\left(x\right)$?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let the velocity of projectile be $u$ and projection angle be $\theta$, then
Horizontal velocity $=u\cos \theta$
Velocity along vertical direction $=u\sin \theta -\frac{gx}{u\cos \theta }$

So, velocity of a projectile at any instant of time $\left(t\right)$ is
$v^2=v_x^2+v_y^2=(u\cos \theta {)}^2+{(u\sin \theta -\frac{gx}{u\cos \theta })}^2$
$\Rightarrow v^2=u^2{\cos }^2\theta +u^2{\sin }^2\theta +\frac{g^2{x}^2}{u^2{\cos }^2\theta }-2gx\tan \theta$
$\therefore KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2-mgx\tan \theta +\frac{m{g}^2{x}^2}{u^2{\cos }^2\theta }$
The given equation represents the equation of a parabola.
Note: Answer cannot be (b) as KE of a projectile never becomes zero.