# Homework Help: Answer check: A ball on a spring launched into the air. find speed/height

1. Nov 11, 2012

### joe426

1. The problem statement, all variables and given/known data

2. Relevant equations

Since energy is conserved, MEinital = MEfinal

3. The attempt at a solution
1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2

After finding what equals 0, im left with:
1/2kxi2 = 1/2mvf2 + mgyf
v=5.97m/s

I found yf by: mgyf = 1/2kxi2
yf = 3.63m

Just checking to see if these answers are correct. THank you for the help!!!!

2. Nov 11, 2012

### haruspex

I get a different result. Please post more details of your calculation.

3. Nov 12, 2012

### joe426

First I found yf, which is the answer to part b.
Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with.

mgyf = 1/2kxi2
y= (1/2kxi2) / g
y=3.63m

Then,

1/2kxi2 = 1/2mvf2 + mgyf

sqrt((kxi2 - mgyf) / m) = v
v = 5.97m/s

Last edited: Nov 12, 2012
4. Nov 12, 2012

### haruspex

Compare those two equations. Shouldn't you get vf=0? ...
.. but you didn't because you dropped a factor of 2.
For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.

5. Nov 12, 2012

### joe426

oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air.

So its,
1/2mvi2 + 1/2kxi2 = mgyf
vi = sqrt( 2gyf - kxi2 )

But now I'm stuck not knowing what it's final position is.

6. Nov 12, 2012

### joe426

Using yfinal = 3.63m. I got the velocity to be 7m/s

7. Nov 12, 2012

### joe426

I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring

V= 8.3m/s for part a
Y= 3.6m for part b

Thanks for the help!!!