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Physics- Energy problem?

  1. Sep 19, 2010 #1
    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. jcsd
  3. Sep 19, 2010 #2
    bump :/
     
  4. Sep 19, 2010 #3
    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?
     
  5. Sep 19, 2010 #4
    so the velocity for horizontal is going to stay 26.241 m/s, right?

    Then I just simply subtract the two velocities?
     
  6. Sep 19, 2010 #5
    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.
     
  7. Sep 19, 2010 #6
    Well I subtracted 11.97 (ball1) - 26.241 (ball2) = -14.271. It came out to be wrong

    The answer is -9.37 m/s :/
     
  8. Sep 19, 2010 #7
    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.
     
  9. Sep 19, 2010 #8
    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
  10. Sep 19, 2010 #9
    Your vertical velocity (7.3) for Ball 1 is wrong.
     
  11. Sep 19, 2010 #10
    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 ?
     
  12. Sep 19, 2010 #11
    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
     
  13. Sep 19, 2010 #12
    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 ?
     
  14. Sep 19, 2010 #13
    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
     
  15. Sep 19, 2010 #14
    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?
     
  16. Sep 19, 2010 #15
    Yes, it is 2 dimensional, but Ball 2 experiences only vertical velocity, since it is thrown straight down.
     
  17. Sep 19, 2010 #16
    makes sense. So in that case we can add the two velocities without using Pythagorean ?
     
  18. Sep 19, 2010 #17
    If you did the kinematics method (V^2 = U^2 + 2AD), you should get the final velocity which is after addition already.
     
  19. Sep 19, 2010 #18
    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.
     
  20. Sep 19, 2010 #19
    What is the answer?
     
  21. Sep 19, 2010 #20
    Answer given is -9.37 m/s
     
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