Asked by Anonymous
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?
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?
Answers
Answered by
Anonymous
Also, what is the velocity of the boulder as it hits the ground?
Answered by
bobpursley
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)>
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