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Help with Projectiles shot horizontally problem

  1. Sep 17, 2009 #1
    I am in an introductory physics class and doing one of these problems, and though it probably isn't too difficult it has stumped me, mainly due to the fact that it asks for vi instead of d or something like that. Anyway, I'm just going to paraphrase the question:

    Suppose a basketball player 2.45 m tall throws a basketball at that height and it lands with a speed of 12.0 m/s. What is the ball's initial speed?

    The equations you might have to use are simple DVAT's:

    df=di+vit+.5at^2
    vf=vi+at

    These 2 primarily.

    So,

    d=-2.45
    vf=12.0 m/s (I think?)
    a=-9.80 m/s^2

    From this I got t=.707 s, and I think I should be able to get the vi in the next step or two, but I can't figure out what those steps are. Help please?
     
    Last edited: Sep 17, 2009
  2. jcsd
  3. Sep 17, 2009 #2
    Hi pbdude :smile: Welcome aboard!

    Was the ball thrown horizontally? (I am guessing from the title that it was).

    Since this is projectile motion, you should break up the DVAT equations into both 'x' and 'y' directions.

    If the ball was indeed thrown horizontally, then when it hits the ground it will hit at an angle, correct? So I do not think that we can assume that 12.0 is vf in just the y-direction as you have in your calculation.

    Could you please post the question exactly as it was posed? I think there may be a key word or detail that will make this a lot easier :smile:
     
  4. Sep 17, 2009 #3
    I was wondering If I had gotten vf right. It didn't sound quite right to me.

    Here's the exact problem:

    "Libyan basketball player Suleiman Nashnush was the tallest basketball player ever. His height was 2.45 m. Suppose Nashnush throws a basketball horizontally from a level equal to the top of his head. If the speed of the basketball is 12.0 m/s when it lands, what was the ball's initial speed? (hint: Consider the components of final velocity)"

    By the way, my teacher gives us the answers first so we can make sure we did it right, and the answer to this is 9.79 m/s.
     
  5. Sep 17, 2009 #4
    hello, i have a problem just like that, and i didnt want to start a new thread, so can i post it here?

    anyways mine's like that but the height is 1.5m(a table) and the velocity(ball rolling off of it) is 5m/s...how do i solve for Vx, Vy, and Dx?
     
  6. Sep 17, 2009 #5
    First you have to solve for time using -1.5m as Dy (since the ball is falling down) and using Vy as 0 and Ay equaling -9.80. From there, you should be able to find time using Df=Di+Vit+1/2at^2.

    After finding the time, use d=rt to find Dx.

    I don't know how to find Vx and Vy....
     
  7. Sep 17, 2009 #6
    Okay then :smile: So I was wrong, you paraphrased just fine. The problem is a little challenging because so little info was given, But that's good :wink:

    I am a little tired, so I might not last too long, but let's get you thinking about it.

    Since we know that it was thrown horizontally, what can you say about the initial velocity in the Y-direction? That is, what is (Vo)y?
     
  8. Sep 17, 2009 #7
    sweet, thanks pbdude :P
     
  9. Sep 17, 2009 #8
    closertolost: Welcome to PF! Could you actually please start your own thread for this question? It gets a little too crowded in these things and we want to make sure we are addressing everyone's questions appropriately. Thanks!
     
  10. Sep 17, 2009 #9
    sure, no problem!
     
  11. Sep 17, 2009 #10
    It is obviously 0 m/s, since x and y are independent of each other. So the question wants me to find (V0)x.

    And I know that frictionless horizontal acceleration is 0 so (Vf)x cannot be 12.0 m/s if (v0)x is 9.79 m/s.
     
  12. Sep 17, 2009 #11
    Right. Okay, so from the first kinematic equation you wrote "df=di+vit+.5at^2" we can establish how long the ball was in the air for by applying it in the y-direction.

    Then using your second equation in the y-direction you should be able to find (Vf)y.

    Knowing (Vf)y and Vf (given) you should be able to find (Vf)x.

    But as you told me already, (Vf)x=(Vo)x due to the no horizontal acceleration condition.
     
  13. Sep 17, 2009 #12
    My main, question, I suppose, is, is the question asking me to solve (V0)y or (V0)x? I'm assuming y. But, I also have already assumed that the V initial for y is zero, so what exactly am I solving for?!
     
  14. Sep 17, 2009 #13
    You have already told me!

    So find Voy first and then use that along with the total V (which was given) to find Vox.

    Think 'right triangles' :wink:
     
  15. Sep 17, 2009 #14
    OK, so Vf of y is -6.93 m/s. You said I could use that to find Vf, thus Vi. How would I go about doing that?
     
  16. Sep 17, 2009 #15
    Oh my God! I did it using the Pythagorean theorem! Thank you so much for your help!
     
  17. Sep 17, 2009 #16
    The total final velocity V=12.0 m/s is a vector that is aiming at the ground at some angle, right? So think of this vector as being the 'hypoteneuse' of the right triangle whose 'legs' are Vy and Vx.

    Get it?
     
  18. Sep 17, 2009 #17
    How sweet it is!
     
  19. Sep 17, 2009 #18
    Yeah, I got it. :)

    Thanks again.
     
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