43.A cricket ball is hit for a six leaving the bat at an angle of 45degrees to the horizontal with Kinetic energy 'K'.At the top, Kinetic energy of the ball is
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At the beginning total energy = 1/2mv2=1/2mv2
At top 1/*m*(vcos45)=1/2*m*v^2*1/2
K'/k= 1/2
At top 1/*m*(vcos45)=1/2*m*v^2*1/2
K'/k= 1/2
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Answer:
At the beginning total energy = 1/2mv2=1/2mv2
At top 1/*m*(vcos45)=1/2*m*v^2*1/2
K'/k= 1/2
Explanation:
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