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Projectile motion and inital speed problem

  1. Oct 29, 2006 #1
    Hey, I'm having trouble with this problem:

    A projectile is fired with an initial speed of 29.0 at an angle of 65.0 degrees above the horizontal. The object hits the ground
    8.00s later.

    How much higher or lower is the launch point relative to the point where the projectile hits the ground?
    Express a launch point that is lower than the point where the projectile hits the ground as a negative number.

    To what maximum height above the launch point does the projectile rise?

    So i'm pretty sure i've drawn my diagram right, and now i've tried to set up an equation where:

    distance x = (vcos)t and

    distance y = (vsine)t -.5gt^2

    when i put in the values from the question i get the wrong answers. if anyone could point me in the right direction and tell me waht i'm doing wrong that would be great.

    Thanks!
     
  2. jcsd
  3. Oct 29, 2006 #2

    OlderDan

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    29 what? Are you doing the problem in the correct units? Maybe you are having difficulty because of the way the question is asked. Your equations are OK, but they assume x and y are zero at time zero. The first part of the question is worded such that it makes mores sense to use a y_o term that will make y = 0 at t = 8 sec. The later part is worded such that it makes more sense to do it as you have done. Make sure you are expressing the height of the launch point relative to the landing point in the first part, and the maximum height relative to the launch point for the second part.
     
  4. Oct 29, 2006 #3
    oh sorry it's 29m/s
     
  5. Oct 29, 2006 #4

    OlderDan

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    OK.. your equations are good, so it's either a computation error or you are not interpreting their statements correctly.
     
  6. Oct 29, 2006 #5
    i'm sorry to keep bothering you but honestly i just have no idea whats going on. If you could explain to me that would be really appreciated!

    Thanks
     
  7. Oct 30, 2006 #6

    OlderDan

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    I get that the launch point is 103.3m above the landing point. The maximum height is about 35.2m. That is what your equations should be giving you. What did you get? If it is different from mine, then we need to figure out why.
     
  8. Oct 30, 2006 #7
    well my problem is that i set up the equations and then i don't know what to do with them. I usually just 'plug' in numbers but its not very useful for undersanding
     
  9. Oct 30, 2006 #8

    OlderDan

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    Right you are. In this problem you don't need the x equation at all. You just need to know that the y motion is independent of the x motion. You do need another equation of y motion for the second part. You can use

    0 = Vosinθ - gt to find the time when the vertical velocity is zero (maximum height) and then use that time in your y equation to find y at that time.

    You could also use V^2 = (Vosinθ)^2 - 2gh with V=0 to find the maximum height h.
     
  10. Oct 30, 2006 #9
    well i know how to set up the equations, but then after that i just 'plug' in numbers but i'm not really understanding what is going on
     
  11. Oct 30, 2006 #10

    OlderDan

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    Practice will help. It can also be helpful to work with other people if you are trying to contribute and not just take what they do. It is one thing to watch other people do math or physics. It is quite something else to be doing it.

    If you tell us more completely what you have tried, we can probably better understand what is wrong with your thinking. In this case you could have shown your calculations instead of just writing the equations, or at least posted the answers you got from them.
     
  12. Oct 30, 2006 #11
    okay thank you, that makes it more clear now of what i need to work and improve on and how i can show my steps more clear. Thank you so much for your help and your time, i really appreciated it!!!
     
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