A bady of 2kg falls from rest. What will be its kinetic energy during the fall at the end of 2s? Assume g= 10m/s^2
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m=2kg
u=0
t=2s
g=10m/s^2
than.
firstly we find v
so,v=u+gt
v=0+10*2
v=20m/s
so,
K.E=1/2mv^2
=1/2*2*20^2
=1/2*2*400
=400j
hope it helps u
u=0
t=2s
g=10m/s^2
than.
firstly we find v
so,v=u+gt
v=0+10*2
v=20m/s
so,
K.E=1/2mv^2
=1/2*2*20^2
=1/2*2*400
=400j
hope it helps u
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