Question

A man drops a 10 kg rock from the top of a 5m ladder .What is it's speed just before it hits the ground ? What is the its kinetic energy when it's reaches the ground?

Answer 1

potential energy at the top=m*g*h                           
                                            =10X10X5 (considering g=10m/s)
                                             =500J

kinetic energy when it reaches the ground=1/2mv²
                                                                   =1/2*10*v²

as we know,
potential energy=kinetic energy (law of conservation of energy)
                       500J=500J
∴kinetic energy = 500J

the speed of rock before reaching the ground is 
                             1/2*10*v²=500
                                   5*v²=500
                                      v=500/5
                                          v=10m/s

Answer 2

potential energy at the top=mgh
=10×10×5
(g=10m/s)
=500J
kinetic energy when it reaches the ground=1/2×mv^2
potential energy=kinetic energy(law of conservation of energy)
500J=500J
°•°kinetic energy=500J
the speed of the rock before reaching the ground=1/2×10×v^2=500
=5×v^2=500
v^2=500/5
v^2=100
v=10m/s