Question

(b) An object of mass 1 kg is dropped from a tower of height 500 m. Find the
kinetic energy and potential energy after 2 seconds.
​

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

Explanation:

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