Question

A body of mass 5kg is dropped from a height .What is its kinetic energy at the end of thethird second of its fall​

Answer 1

The time at which body touches ground is t=

2d/g

=

(2∗500)/10

=10s

Therefore v=gt=10∗3=30m/s

Therefore KE=

2

1

∗m∗v

2

=

2

1

∗3kg∗30

2

=1350J