A ball is dropped from 27.0 ft above the ground. Using energy considerations only, what is the velocity of the ball just as it hits the ground? If the ball is thrown back up to a ledge 15 m above the ground, what would be the velocity just as it reaches the ledge?
2 answers
Help with physics queston.
the potential energy at the start is mgh
That should equal the kinetic energy 1/2 mv^2 at the bottom.
On the last, assuming it was thrown up to reach 27ft, then
original energy=newpeat15 + KE at 15
mg*27=mg15+ 1/2 m v^2
That should equal the kinetic energy 1/2 mv^2 at the bottom.
On the last, assuming it was thrown up to reach 27ft, then
original energy=newpeat15 + KE at 15
mg*27=mg15+ 1/2 m v^2