Question

a body is projected from ground at an angle theta 1 with horizontal with kinetic energy k. what is the potential energy of the body when velocity makes angle theta2 with horizontal

Answer 1

Let a body of mass m is projected from ground with speed u m/s at an angle $\theta_1$ with horizontal with kinetic energy k.

so, initial kinetic energy = 1/2 mu²

k = 1/2 mu² ......... (1)

we know, horizontal component of speed in case of projectile always remains same.

e.g., horizontal component of initial speed = horizontal component of final speed

or, $ucos\theta_1=vcos\theta_2$

or, $v=\frac{ucos\theta_1}{cos\theta_2}$

now, kinetic energy at that point = 1/2 mv²

= 1/2 m$\frac{u^2cos^2\theta_1}{cos^2\theta_2}$......(2)

from energy conservation theorem,

$K.E_i=K.E_f+P.E$

or, $\frac{1}{2}mu^2-\frac{1}{2}m\frac{u^2cos^2\theta_1}{cos^2\theta_2}=P.E$

or, $P.E=\frac{1}{2}mu^2\left[1-\frac{cos^2\theta_1}{cos^2\theta_2}\right]$

from equation (1),
$P.E=k\left[1-\frac{cos^2\theta_1}{cos^2\theta_2}\right]$
hence, potential energy of the body when velocity makes angle $\theta_2$ with horizontal is $P.E=k\left[1-\frac{cos^2\theta_1}{cos^2\theta_2}\right]$