Question

If a body of mass 3kg is dropped from the top of a tower of height 25m then it's k.E after 3s?

Answer 1

Answer:

1350 joule

Explanation:

velocity after 3 seconds

v=u+at

v= 10×3

v= 30m/s

now,

$k.e = \frac{1}{2} m {v}^{2}$

so, k.e=1/2 ×3 ×(30)^2

k.e= 1350 joule.

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