Question

A ball is thrown at an angle 45° to the horizontal with kinetic energy K what is the kinetic energy at the highest point of the trajectory​

Answer 1

Answer:

• The velocity of ball at highest position is equal to ucos teta
• Kinetic energy = 1/2 mv square

= 1/2 m( ucos45°)2

=1/2 mu(1/√2)2

=1/2 mu2×1/2

(1/2 mu2= k)

=K/2.

The answer is k/2