Question

- A Body of mass long is let loose from
a tower op 10om height. Calcellate the
I kinetic energy of the body when it is
at the bottom of the tower.
​

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