when a ball comes to ground in 3 sec of mass 101g .what will be the kinetic energy of ball?
yashu63:
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If it is provided that ball was release at height in rest position, then by kinematical equation,
V=U+at
Here V is to be calculated where U=0(rest) and a=g(acceleration due to gravity), t=3sec
V=0+9.8×3
V=29.4m/s
Now as we know Kinetic energy
KE=½MV²=½×0.101×29.4........(taking mass in SI unit)
KE=1.484 Joules
Note that if it is not provided that ball was first at rest then the initial velocity of the ball should be taken as U and the final velocity V is calculated. The process will be the same but the answer would differ in that case! Thank you.
V=U+at
Here V is to be calculated where U=0(rest) and a=g(acceleration due to gravity), t=3sec
V=0+9.8×3
V=29.4m/s
Now as we know Kinetic energy
KE=½MV²=½×0.101×29.4........(taking mass in SI unit)
KE=1.484 Joules
Note that if it is not provided that ball was first at rest then the initial velocity of the ball should be taken as U and the final velocity V is calculated. The process will be the same but the answer would differ in that case! Thank you.
it will also be easy to take the gravity as 10m/s square
the answer will come 0.101 ×10×3=3.03
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