Question

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

Answer 1

as we know that when there is a free fall of an object,the decrease in potential energy at any point on its path,appears as an equal amount of increase in its kinetic energy ...because by the law of conservation of energy energy can neither be destroyed nor be created.

mechanical energy=kinetic +potential
 (mass=10kg,g=10m/s^2,h=5m)
potential energy at the top(maximum)=mgh                           
                                                           =10X10X5
                                                           =500J
kinetic energy when it reaches the ground(maximum)=1/2mv^2
                                                                                     =1/2*10*v^2
     potential energy=kinetic energy            (law of conservation of energy)
                      500J=500J
                              
 therefore,kinetic energy=500J
                THE SPEED JUST BEFORE IT HITS THE GROUND  IS-
                             1/2*10*V^2=500
                                           V^2=500/5
                                          V=10
THEREFORE V=10                                                                                   

Answer 2

Mass = 10 kg
Height = 5 m
so, by equation of position velocity relation
v² - u² = 2as
u = 0 m/s
a = g =10 m/s²
s = h = 5m
so, velocity just before hitting the ground = 10 m/s
and the kinetic energy = 1/2 mv² = 500 J