A 2kg stone is droped from the top of a 20m building

Question

a)what potential energy does it posses
b)At what height does its potential energy become equal to its kinetic energy?
c)what is its kinetic energy just before it hits the ground?
d)With what velocity does it reach the ground?
2)what does the velocity law of conservation of energy states?
3)Define kinetic energy?

Answer 1

PE = mgh
when half the PE is gone!
same as the original PE
set mgh = 1/2 mv^2 and solve for v

Answer 2

P:E 400J

Answer 3

Kindly help me with the question above

Answer 4

ANSWER

Answer 5

KE=1/2 mv2
g=10m/s2
KE=O
PE =mgh
PE =2*10*20
PE=400jous