Question

A ball whose K.E is E is projected at an angle of 45 degree to the horizontal the K.E of the ball at the highest point of its flight will be

Answer 1

Kinetic Energy, E = ½mv2

At the highest point, vertical component of velocity becomes zero and only the horizontal component is left that is given by

ux = u cosθ

θ = 45°, so cosθ = 1/√2

ux = u/√2

Kinetic energy E at the highest point = ½m(u/√2)2 = E/2