a ball (250 g) thrown straight up into the air. reaches an altitude of 15 m then returs to you.?

Question

what is the gravitational PE of the ball at its highest position?
what is the KE as soon as it leaves your hand?

Answer 1

That is a very neat application thought as well as an analytical problem.

First, since the ball reaches the top most point by virtue of the energy it possed when it left your hand, both must be the same. only that they were in different forms.
Second, we know that
V = U + 2*A*S
where V is the final, U is the initital velocity, A is acceleration due to gravity and S is the distance.

Since the direction is against gravity, A = -9.8m/s^2
The final velocity when the ball is at the apex, is zero before it starts falling again.

Now, the equation is
0 = U -2*9.8*15
=> U = 294 m/s

We also know that Kinetic energy is given by
KE = 0.5*M*Velocity^2
= 0.5 X 0.250 X 294^2
= 10804.5 Kgm/s^2

This must be equal to the potential energy, by our first discussion. Just give it the opposite sign to denote that it acts in the opposite direction.

Answer 2

Aww, darn.... I computed force, not energy.

I like the PE = mgh equation above, but the KE and the PE should be the same number, since the ball had NO PE when it was released, all the KE should be stored as PE.

PE at top = KE when released = 0.250 * 10 * 15 = 37.5 J

Answer 3

PE= mgh = (0.25kg) * (9.8 m/s2) * 15 = 36.65 J
KE = 0.5*m*v^2 => 36.65 = 0.5*0.25*v^2 => v^2 = 293.2 => v = sqrt of (293.2) units in m/s.