Question

A ball whose kinetic energy E, is
projected at an angle of 45° to the
horizontal. The kinetic energy of the
ball at the highest point of its height will
be
E
e/v2
E/2
zero​

Answer 1

Answer:

The kinetic energy of the ball at the highest point is as follows:

the expression for kinetic energy is KE=21​mv2.

So, the kinetic energy at the highest point is,

KEh​=21​m(vcos(45∘))2

KEh​=21​m(v2)(cos(45∘))2

KEh​=E(cos(45∘))2

KEh​=2E​

so, at the highest point the kinetic energy is 2E​.

Explanation:

Answer 2

answer: hope it really helps you...