A boulder with a mass of 48.3kg is sitting on the top of a 11.8m tall cliff, and the boulder falls off of the cliff. At what height above the ground is the boulder when it reaches a velocity of 6.40m/s?
I found that the kinetic energy of the boulder before it falls off is 5585.412 Joules.
I also found out that when the boulder reaches the velocity of 6.40m/s, its kinetic energy is 989.184 Joules.
Are these correct so far?
2 answers
Also, what is the velocity of the boulder as it hits the ground?
velocity at ground:
InitialPE=finalKE
mgh=1/2 m v^2
v=sqrt(2gh)=sqrt(2*9.8*11.8)
notice you wrote initial KE, you meant initial Potential energy. The initial KE is zero (sitting on a cliff)>
InitialPE=finalKE
mgh=1/2 m v^2
v=sqrt(2gh)=sqrt(2*9.8*11.8)
notice you wrote initial KE, you meant initial Potential energy. The initial KE is zero (sitting on a cliff)>
4 answers