A particle is projected making an angle of 45∘ with horizontal, with kinetic energy K. The kinetic energy at highest point will be
A
K√2
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B
K2
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C
2K
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D
K
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Solution
The correct option is BK2
Kinetic energy at initial point: K1=12mv2...(1)
Kinetic energy at highest point: K2=12mv′2Velocity at highest point,v′=v√2K2=12m[v√2]2=14mv2...(2)
On comparing eq.(1) and eq. (2), K2=12K1
or K2=12K