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Need help for a projectile problem

  1. Sep 27, 2004 #1

    mad

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    Hello..
    There is a projectile problem that I can't seem to resolve.. I have tried it for more than 30min now and can't find out how to do it.. I got an exam in 2 days so would someone please help me and tell me how to do this exercice:

    A basket-ball is launched at an angle of 45 degres (it is on the ground). The basket is at an horizontal distance of 4m and a vertical distance of 0,8m. What is the initial velocity (Vo) required to reach the basket?

    .. sorry for my english this is translated from french.. here is what I tried to do but I know its not good..

    http://img39.exs.cx/img39/4460/gdg.jpg

    thank you
     
    Last edited: Sep 27, 2004
  2. jcsd
  3. Sep 27, 2004 #2

    Tide

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    First, set [itex]y_0 = 0[/tex] because they gave the height of the hoop relative to the starting height.

    You should be able to combine your first two equations to give y as a function of x and when you solve that equation for [itex]v_0[/itex] you will find
    [tex]v_0 = \sqrt \frac {g x^2}{x-y}[/tex]
    which actually works out to a nice round number in your problem!
     
  4. Sep 27, 2004 #3

    mad

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    Thanks a lot.. but would you explain me how you got that formula and what is wrong with what I did so I can learn to not do it again?

    Thank you very much for your time
     
  5. Sep 27, 2004 #4

    Tide

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    Since the launch angle is 45 degrees
    [tex]y = \frac {v_0 t}{\sqrt 2} - \frac {1}{2} g t^2[/tex]
    [tex]x = \frac {v_0 t}{\sqrt 2}[/tex]
    Solve the second equation for t and substitute into the first:
    [tex]y = x - \frac {g x^2}{v_0^2}[/tex]
    from which my earlier equation follows.
     
  6. Sep 27, 2004 #5

    Tide

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    P.s.

    As to what you did wrong I think you just got bogged down in your algebra.
     
    Last edited: Sep 27, 2004
  7. Sep 27, 2004 #6

    mad

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    I feel very dumb for asking this.. but where does [tex]{\sqrt 2}[/tex] come from?

    I know that X=Vxo*t which gives X=Vo * Cos45* t
    and Y= Yo +Vyo*t - 1/2 *g * t^2

    but why do you put a [tex]{\sqrt 2}[/tex] ?

    Thanks very much for your time
     
  8. Sep 27, 2004 #7
    [tex]\cos 45^\circ=\sin 45^\circ=\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}[/tex]
     
  9. Sep 27, 2004 #8

    mad

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    YES!!! I finally got it! I was doing it alright but I just messed up with my algebra like you said.. except you gave me a much simpler way. I just couldnt use the [tex]{\sqrt 2}[/tex] because my teacher never told use about that.

    Also, when do I have to use this: 2 sin A * cos A = Sin 2A ? In what type of problems is this used?
    thanks again
     
  10. Sep 27, 2004 #9

    Tide

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    The double angle shows up in trajectory type problems when you try to calculate the range of a projectile. We didn't have to invoke it for the problem you posed.
     
  11. Sep 27, 2004 #10

    mad

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    I was doing another problem and I had to use this. It was
    (9.8 * 53) / 2* 87^2 = sinX* cosX

    so I did this:
    (9.8 * 53)/ 87^2 = 2*sinX*cosX
    -->
    (9.8 * 53)/ 87^2 = Sin 2X

    but I have no idea how to find this.. the teacher never talked about it. Ill try to find out on the internet. I just want to thank you again for all those replies and help.
     
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