Question

An object of mass 3kg was released from 40m above the ground. Using the conservation of energy, determine the velocity of the object right before hitting the ground.

Answer 1

Answer:

v=28m/s

Explanation:

Conservation of Energy

Ei=Ef

mgh= 1/2mv^2

9.8*40*2= v^2

784= v^2

v= 28m/s