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A particle is projected making angle 45^...

A particle is projected making angle `45^@` with horizontal having kinetic energy K. The kinetic energy at highest point will be : -

A

zero

B

`k`

C

`(k)/(sqrt(2))`

D

`(k)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
At the top of trajectory,
`K'=(1)/(2)m(u cos theta)^(2)`
`=(1)/(2)mu^(2).cos^(2)45^(@)=(k)/(2).`
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