Question

A particle is projected at 60° to the horizontal woth keintic energy K what is the kinetic energy at highst point

Answer 1

let initial velocity be = u
u=ucosθ i + usinθ j
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=> kinetic energy at the Point Of Projection =K

=>K= (1/2) m(ucosθi+usinθj)^2
=>K=(1/2) m [√(u^2cos^2θ + u^2sin^2θ)^2]
=>K=(1/2)mu^2
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we know that on HIGHEST POINT

uy=0 and ux= ucosθ

=>K'=(1/2)m(ucosθ)^2

=>K'=(1/2) m (ucos60°)^2

=>K'=(1/2) m (u/2)^2

=>K' = [(1/2)mu^2]/4

=>K'=K/4
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hope it helps you....
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