Finding Velocity of a dropped ball

[cowmoo32]

A baseball of mass .025kg is released at rest from a location 3.5m above the ground. You will calculate its speed at the instant it hits the ground.

Choose a sysem to analyze, and list the objects in your system.
I chose the baseball and the Earth as the system

Write out the energy principle as it applies to the system you have chosen. Solve for the speed of the baseball when it hits the ground
Here's where I'm lost. The baseball has kinetic energy (1/2mv^2), and the Earth is pulling down on it (-9.8*m), but I'm not sure how to arrive at the answer. According to the key, the answer is 8.3m/s.

 

[berkeman]

If you are supposed to solve this problem using conservation of energy, you need to know the potential energy (PE) of the ball before it is dropped. How much energy did it take to initially lift the ball from the ground to the height h?

 

[cowmoo32]

Wouldnt that be M*G*D? .025*9.8*3.5 = .8575

 

[berkeman]

Yep. And what is the total energy (TE) of the system? What can you say about it? How does that help you solve for the ball's final velocity?

 

[cowmoo32]

Total energy has to equal zero.

K+U+G=0
(.5mv^2)+.8575-(9.8*m)=0
correct?
or would it be
K+G=U?

 

[berkeman]

I'm guessing that you mean KE for K, but what are U and G? I use TE=KE+PE myself.

 

[cowmoo32]

oh, G=gravity U=potential energy K=kinetic energy

 

[berkeman]

oh, G=gravity U=potential energy K=kinetic energy

Gravity is not an energy. Different units, different concept. All the energies in the equation should have units of joules.

 

[cowmoo32]

ahhh...ok.I'm still not sure how to find velocity though. If I have KE+PE=TE, I'm still left with 2 variables. Because my velocity is unknown for KE, then I don't have a number for TE.

 

[berkeman]

moo, the concept here is that the TE does not change when you drop a ball. What is lost from the initial PE is gained in the KE. What is the initial PE from lifting the ball from the ground to a height h? Be sure to include units for all quantities in that equation that you will write. Then assuming that all that PE is lost and converted to KE in the fall, what is the final KE? What does that mean the final v is? I got to go now. You're almost there.

 

[fffff]

PE of the ball is equal to the KE of the ball as it hits the ground

 

[berkeman]

PE of the ball is equal to the KE of the ball as it hits the ground

That's not stated very well. More accurately, the TE of the ball is all PE when the velocity is zero (at the top), and the TE is all KE just before it hits the ground.