Question

A particle is projected at 60° to the horizontal with a kinetic energy K' the kinetic energy at highest point is

Answer 1

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

The body Is projected with a velocity "u."

The body Is projected with a velocity "u."and the kinetic energy is K'

Kinetic energy at topmost point be K.

$\huge{\underline{Explanation:-}}$

• At the time of projection

$K' = \frac{1}{2} m {u}^{2} \: \: - - (1)$

At highest point vertical component is Zero.

${u_y}$ = 0.

Horizontal component remains constant throughout the journey.

${u_x}$ = u cos ${\theta}$

So, Kinetic energy becomes

$K = \frac{1}{2} m {(u \cos \theta)}^{2}$

equation becomes

$K = \frac{1}{2} m {u}^{2} { \cos( \theta) }^{2}$

From equation 1.

$K = K' \ {cos( \theta) }^{2} \:$

As ${\theta}$ = 60°.

as cos 60 is 1/2.

$K = K' \times \frac{1}{4}$

$K = \frac{K'}{4} \:$

$\boxed{\boxed{K = \frac{K'}{4}}}$

So,the kinetic energy is 1/4 times from the initial Kinetic energy.