A cricket ball of mass M is hitted at the angle 45 degree to the horizontal with velocity V. It's kinetic energy at the topmost point is
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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
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
rohit2875:
am I correct
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