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Trajectory Problem

  1. Sep 16, 2008 #1
    I'm having difficulty figuring out this problem.
    Is there a formula I can use?
    A juggler throws a ball straight up into the air with a speed of 14m/s. With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?
     
  2. jcsd
  3. Sep 16, 2008 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    What equations do you think apply?

    For instance what formulas do you know that might relate velocity to height? How long does it take the first ball to get there?

    If you know how long you have and you wait another .5 seconds then how fast does the second ball need to be to get to that height at that same time?
     
  4. Sep 16, 2008 #3
    1. The problem statement, all variables and given/known data
    I'm having difficulty figuring out this problem.
    Is there a formula I can use?
    A juggler throws a ball straight up into the air with a speed of 14m/s. With what speed would she need to throw a second ball half a second later, starting from the same position as the first, in order to hit the first ball at the top of its trajectory?

    I'm pretty sure that I am supposed to use acceleration as -10m/s^2, so I think I need to know how long it would take for the ball to go up and then land back down to find the time. I tried using this formula v(final) = v(initial) + acceleration(change of time), which is 0-14/-10=1.4s. Then I get stuck...I don't know how to get from that information to finding the velocity of the second.
     
  5. Sep 16, 2008 #4
    You should have been given the basic formula in class. You are given the initial belocity, final velocity and acceleration (a) (due to gravity) so you should be able to work out the time (t) for all this to happen. Then you will have t, a and v for the second ball, therefore...
     
  6. Sep 16, 2008 #5

    CompuChip

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    So you can use v(final) = v(initial) + acceleration (time) to find how long it takes to reach the highest point (question: is it 0.7s or 1.4s?)

    Then there is this other formula:
    (travelled distance) = 1/2 * (acceleration) * (time)^2 + (inital velocity) * time + (initial distance)
    which you could use for the second ball
     
  7. Sep 16, 2008 #6
    Ok, so would that mean that the second ball would have the equation:
    traveled distance = (1/2)(10)*(.7^2) + (14)(.7), which equals 12.25m. So then would I divide the distance traveled by 0.7 to get the speed? 12.25/0.7= 17.5m/s. Is this correct?
     
  8. Sep 17, 2008 #7
    Where are you getting 0.7? The problem states "half a second later", not "half the time later".
     
  9. Sep 17, 2008 #8
    Sorry I forgot to ask also -- where are you getting 14? That was the initial velocity of ball 1. The initial velocity of ball 2 is unknown -- it's what you're supposed to be solving for (^:

    The distance traveled is known--or can be determined by plugging ball 1 into

    d = 1/2 (v_initial - v_final) * t
     
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