Homework Help: Physics- Energy problem?

1. Sep 19, 2010

J.live

1. The problem statement, all variables and given/known data
Two baseballs are thrown off the top of a building that is 7.32 m high. Both are thrown with initial speed of 58.7 mph. Ball 1 is thrown horizontally, and ball 2 is thrown straight down. What is the difference in the speeds of the two balls when they touch the ground? (Neglect air resistance.)

vball 1 - vball 2 = ?

2. Relevant equations

Ei = Ef

3. The attempt at a solution

58.7 mph -> 26.08 m/s ? (not sure)

I used

Ei= Ef

Ball 1 ( x - direction)

Ki (0) + Ui (mgh)= 1/2mv^2+ Uf(0)

gh= 1/2 V^2

V= 11.97m/s ( same velocity for Ball 2 (y-direction))

I just dont know how to find the difference. If someone can explain it in steps. Thanks

2. Sep 19, 2010

J.live

bump :/

3. Sep 19, 2010

SiYuan

58.7 mph = 26.241m/s
This is not an energy problem per se, more of a kinematic one.

Since air resistance is negligible, you can assume that the horizontal velocity of the ball thrown horizontal to be constant,

Does that help now?

4. Sep 19, 2010

J.live

so the velocity for horizontal is going to stay 26.241 m/s, right?

Then I just simply subtract the two velocities?

5. Sep 19, 2010

SiYuan

Yep. Since the question is asking for the difference in speed, as Ball 1 falls to the ground it gains vertical velocity as well (how else would it fall?), and Ball 2 will fall to the ground, gaining vertical velocity.

6. Sep 19, 2010

J.live

Well I subtracted 11.97 (ball1) - 26.241 (ball2) = -14.271. It came out to be wrong

The answer is -9.37 m/s :/

7. Sep 19, 2010

SiYuan

No no, that's not quite how you do it.

Let's keep the horizontal velocity of both balls out of the picture for now.

The vertical velocity of Ball 1 will increase from 0 to X as it falls to the ground,

whilst the vertical velocity of Ball 2 will increase from 26.24(thrown to the floor) to (26.24 + Y) as it falls to the ground. Calculating it is simply, v^2 = u^2 + 2AD where A is acceleration due to free fall and D is the height of the building.

After which, we look at horizontal velocity.

Since the horiz velocity of ball 2 is 0, and will likely stay that way.

The horiz of Ball 1 is constant, which is 26.241.

Hence, just using Pythagoras' on Ball 1, you can find the speed for Ball 1.

The speed of Ball 2 would be the vertical velocity only.

8. Sep 19, 2010

J.live

what the? We have to find the vertical speed of ball 1? Why do we have to do that?
so using Pythagorean

(7.3)^2 + (26.241)^2 = c^2 ?

I am not getting this.

Last edited: Sep 19, 2010
9. Sep 19, 2010

SiYuan

Your vertical velocity (7.3) for Ball 1 is wrong.

10. Sep 19, 2010

J.live

No, I am not to able to understand what we are looking for here? Why do we need to find the vertical velocity for ball 1 in the first place? If so how do we do it? using Pythagorean ?

11. Sep 19, 2010

SiYuan

To find the vertical velocity for Ball 1, you engage the same method as Ball 2, where v^2 = u^2 + 2AD,

Since initial vertical velocity of Ball 1 = 0, hence V^2 = 2AD, V = 11.98m/s

The final velocity of Ball 1 would be C^2 = 11.98^2 + 26.241^2

12. Sep 19, 2010

J.live

oh, so that's how we find the overall velocity for Ball 1 , right?

Then for Ball 2? We have to find the overall velocity for that too?

How would we do that ?

13. Sep 19, 2010

SiYuan

Yep, as I mentioned earlier, the velocity for Ball 2 would be the initial velocity + the additional vertical velocity, which is found by V^2 = U^2 + 2AD as well

14. Sep 19, 2010

J.live

I got 11.98 m/s which is the final velocity for ball 2 as well?

I dont understand how can we simply add 26.41 to 11.98 for ball 2? Isn't it two dimensional like Ball 1?

15. Sep 19, 2010

SiYuan

Yes, it is 2 dimensional, but Ball 2 experiences only vertical velocity, since it is thrown straight down.

16. Sep 19, 2010

J.live

makes sense. So in that case we can add the two velocities without using Pythagorean ?

17. Sep 19, 2010

SiYuan

If you did the kinematics method (V^2 = U^2 + 2AD), you should get the final velocity which is after addition already.

18. Sep 19, 2010

J.live

For Ball 1 11.98^2 + 26.241^2= c^2 ---> 28.84

For Ball 2 using the same method V^2 = U^2 + 2AD
V^2= 26.241^2 + 2 (9.8)(7.32)
V = 28.84 m/s

I still didn't get the right answer.

What am I doing wrong?

Sorry for bothering you time and again.

19. Sep 19, 2010

SiYuan

20. Sep 19, 2010

J.live

21. Sep 19, 2010

SiYuan

Have you missed out anything from the question?

22. Sep 19, 2010

J.live

No, I have not missed out since I copy pasted it.

What the professor did was use the Kinetic energy equation

Ball 1
Eiy = Efy

Ki + Vi = Kf + Vf

0+ mgh = 1/2 mvf^2+0

Vfy= 11.98 m/s
================
Ball 2

Eiy= Efy

0+mgh= mvf^2 + 0

Vfy= 11.98 m/s
============================
Then he used Pythagorean Theorem for Ball 1

11.98^2 + 26.241^2= C^2

But for Ball 2 He simply added Viy(26.241)+ VfY(11.98)

Which does come out to be -9.38 m/s :/

Last edited: Sep 19, 2010
23. Sep 19, 2010

SiYuan

The initial KE of Ball 2 should not be 0, but instead 1/2 (m)(26.241)^2.

And after calculation, the final velocity still gives me the same answer I had during kinematics.

Your professor has made the mistake that (Vf + Vi)^2 = Vf^2 + Vi^2

24. Sep 19, 2010

J.live

This is strange. Anyways, thanks for your help.