Question

If a body of mass 3 kg is dropped from top of a tower of height 250 m, then its kinetic energy after 3 sec will be​

Answer 1

Answer:

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