A cricket ball is hit at an angle 45° to the horizontal with kinetic energy K the kinetic energy at the highest point is what?
a)k
b)k/2
c)ksin 45°
d)zero
Kinetic Energy, K= ½mv^2
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 K at the highest point = ½m(u/√2)^2 = (1/2 ×(mu^2))/2
=K/2