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Question

An object has kinetic energy E when it is projected at an angle of maximum range then at the highest point of its path the kinetic energy E of the object will be


Solution

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° (Since at maximum range angle will be 45)
, so cosθ = 1/√2

ux = u/√2

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


Physics

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