Question

A particle is projected at 45 degree to the horizontal with a kinetic energy k .The kinetic energy at the highest point is

Answer 1

Kinetic energy of the particle = k

Let the speed with which it is projected be v

let the mass be m

$v = \sqrt{\frac{2k}{m} }$

Now, the velocity at the highest point = v Cos 45°

Hence, the kinetic energy at the highest point:

$k_h = \frac{1}{2} m \times \frac{2k}{m} \times \frac{1}{\sqrt{2^2} }$

$k_h = \frac{k}{2} }$

Hence, the kinetic energy at the highest kinetic energy is: $k_h = \frac{k}{2} }$