### JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

### AIEEE 2002

A ball whose kinetic energy E, is projected at an angle of $45^\circ$ to the horizontal. The kinetic energy of the ball at the highest point of its height will be
A
E
B
${E \over {\sqrt 2 }}$
C
${E \over 2}$
D
zero

## Explanation

Assume the ball of mass m is projected with a speed u. Then the kinetic energy(E) at the point of projection = ${1 \over 2}m{u^2}$

At highest point of flight only horizontal component of velocity $u\cos \theta$ present as at highest point vertical component of velocity is = 0.

Note : The horizontal component of velocity does not change in entire projectile motion.

At highest point the velocity is = $u\cos \theta$ = $u\cos 45^\circ$ = ${u \over {\sqrt 2 }}$

$\therefore$ The kinetic energy at the height point = ${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$

= ${1 \over 2}m{u^2} \times {1 \over 2}$ = ${E \over 2}$