Question

A body is projected at an angle theta to the horizontal with kinetic energy e neglecting air friction the potential energy at the top most point of the path is

Answer 1

Hope it will help you....

Answer 2

Answer:

$P.E.\ =\ esin^2\theta$

Explanation:

Given,

• Angle of projection = $\theta$
• Initial kinetic energy of the body = $K.E.\ =\ e$

Let u be the initial speed of the body

Horizontal initial velocity of the body = $v_x\ =\ ucos\theta$

Vertical initial velocity of the body = $v_y\ =\ usin\theta$

$\therefore K.E.\ =\ \dfrac{1}{2}mu^2\\\Rightarrow  e\ =\ \dfrac{1}{2}mu^2$

There is no effect of the air friction on the bod, therefore at the topmost point of the path the potential energy of the body is equal to the difference in the kinetic energy of the ball and at the top of the path.

The total initial kinetic energy of the ball = kinetic energy + potential energy of the ball at the top of the path.

$\therefore K.E_i\ =\ K.E_f\ +\ P.E.\\\Rightarrow P.E.\ =\ K.E_i\ -\ K.E_f\\\Rightarrow P.E.\ =\ \dfrac{1}{2}mu^2\ -\ \dfrac{1}{2}mu^2cos^2\theta\\\Rightarrow P.E.\ =\ \dfrac{mu^2(1\ -\ cos^2\theta}{2}\\\Rightarrow P.E.\ =\ \dfrac{mu^2sin^2\theta}{2}\\\Rightarrow P.E.\ =\ esin^2\theta$

Hence the potential energy at the top of the path is $esin^2\theta$