Question

a ball whose kinetic energy is E, is projected at an angle 45 degrees to the horizontal. Find the kinetic energy of the ball at the highest point of its flight.

Answer 1

Answer:

hope it will help you.

see the attachment

Answer 2

Answer:

The kinetic energy of the particle will be $1 / 2 \mathrm{E}.$

Explanation:

We have the values given as under

Let us assume the initial velocity of the particle be “v”

Kinetic energy = $1 / 2 m \mathrm{V}^{2}=E-\rightarrow 1$

Particle velocity at the highest point it reaches is = v cos 45

The particle’s kinetic energy at highest point is = $1 / 2 m(v \cos 45)^{2}$

$=1 / 2\left(\mathrm{mv}^{2}\right)(1 / 2)${from equation 1 we can conclude}

$=1 / 2 \mathrm{E}$