A body of 2 kg falls from rest .What will be it's kinetic energy when it just reaches the ground ?*( g= 9.8m/s ^2)
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K.E = 1/2 mv^2
v= velocity before reaching the ground
=√2gh
h=1/2gt^2
=4.9t^2
v=√2(9.8)(4.9t^2)
=√96.04t^2
= 9.8t
K.E = 1/2 (2)(9.8t)
= 9.8t
v= velocity before reaching the ground
=√2gh
h=1/2gt^2
=4.9t^2
v=√2(9.8)(4.9t^2)
=√96.04t^2
= 9.8t
K.E = 1/2 (2)(9.8t)
= 9.8t
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