A boulder with mass 2500 kg falls from a ledge 200 m above the ground. If mechanical energy is conserved what is the speed of the boulder just before it hits the ground?
As mechanical energy is conserved the total energy in the boulder which is the sum of the potential and the kinetic energy is the same throughout its motion.
On the ledge 200 m above the ground, the boulder has no kinetic energy and its potential energy is given as m*g*h = 2500*9.8*200 = 4900 kJ
When the boulder falls from the ledge, its potential energy is gradually converted to kinetic energy. The kinetic energy of the boulder when it is traveling at a velocity v is given by (1/2)*m*v^2.
Just before the boulder strikes the ground all the potential energy of 4900 kJ that it had when it was placed on the ledge has been converted to kinetic energy. If its velocity is v, (1/2)*m*v^2 = 4900000
=> (1/2)*2500*v^2 = 4900000
=> v^2 = 3920
=> v = 62.6 m/s
The velocity of the boulder before it strikes the ground is 62.6 m/s