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Finding Distance Up (Max Height) trouble

  1. Jan 29, 2008 #1
    Hey guys, I just started a physics class in college and I'm having a hard time understanding most of it, but I'm trying...

    Anyways, I'm trying to find "Distance Up" for the problem below, and I just can't understand how the professor got 1.28m for his answer.

    1. The problem statement, all variables and given/known data

    A projectile launched at an angle of 30 degrees, and a velocity of 10 m/s.

    Horiz. Vel. (I got 8.66 m/s)
    Vert. Vel. (I got 5 m/s)
    Time Up (He gave us this, .51 sec)
    Total Time (I got 1.02 sec)
    Distance Up (Can't figure this out...)
    Distance (I got 8.84 m)

    2. Relevant equations

    Distance Up = (Final Vel^2 - Initial Vel^2) / 2 * Acceleration
    or = Initial Vel * Time + 1/2 * Acceleration * Time^2

    3. The attempt at a solution

    Well, I'm having a hard time figuring out how to get Final Velocity with this type of problem....

    I mean, the initial velocity should 0, so when I plug that into the 2nd formula I get:
    (0 * 1.02) + (1/2) * 9.8 * 1.02^2 = 5.09796

    But the correct answer should be 1.28 m....
    Last edited: Jan 29, 2008
  2. jcsd
  3. Jan 29, 2008 #2
    Hi Mirth. Welcome to PF!

    No, it is not. It is given in the problem that the velocity of the projectile at launch is 100m/s. (This is the initial velocity.)

    As for the final velocity, it's the same as the velocity of a ball (that you throw straight up) when it reaches the maximum height. (The ball you throw up is also an example of a projectile, but it doesn't travel horizontally like the object in the problem.)
    Last edited: Jan 29, 2008
  4. Jan 29, 2008 #3
    Oh, I see I see.

    I also messed up in my original post putting 100m/s for velocity, when it's supposed to be 10 m/s. I'll edit that now.

    But anyways, so if the initial velocity is 10 m/s, then in the second formula, it should be:

    10 * 1.02 + (1/2) * 9.8 * 1.02^2

    And I get 15.29796 for an answer, so I'm not sure what I'm putting in wrong. :(
  5. Jan 29, 2008 #4
    I made a slight but significant error in my earlier post, so please forgive me. :) :redface:

    The initial velocity in this case is the vertical component of the velocity with which it was thrown.

    And for the time, you shouldn't be using 1.02s. Can you see why?
  6. Jan 29, 2008 #5
    Hehe, it's no problem.

    So you're saying Initial Velocity is the Vertical Velocity that I found, which is 5 m/s.

    As for the time part... I guess I should be using the "Time Up" I got before (.51 sec) because the point of the highest distance is at the time it took to get up to that point, which is the halfway point...

    If that is so though, then the formula would be: 5 * .51 + (1/2) * 9.8 * .51^2, which I get 3.82449, which is still wrong.

    I'm really terrible at the thinking part of this. :(
  7. Jan 29, 2008 #6

    Correct again.

    What you have there is really a vector equation. (Have you studies those yet?) It takes into account the directions of certain quantities as well. As per convention, in an x-y coordinate system, the +y axis points "upwards." The initial velocity points up and is therefore positive, but what about gravity?
  8. Jan 29, 2008 #7
    We haven't really learned anything about this chapter yet in class, so I'm just trying to use my book as a source (it feels horribly unorganized...), so I guess we haven't studied vector equations yet.

    As for your question, I guess gravity would be negative because it's is pulling the object down...
  9. Jan 29, 2008 #8
    And how would that affect the equation

  10. Jan 29, 2008 #9
    It'd be a negative number; -9.8

    That's the formula he gave us to find Distance for this problem. Even if I use the -9.8, I get a really odd number.

    I think I'm just gonna skip this problem.
  11. Jan 29, 2008 #10

    (0.5*0.51) - (0.5*9.8*(0.51)^2) = Answer
  12. Jan 29, 2008 #11
    Hrm, yeah, I've never seen that formula, and that's not what he gave us so that totally threw me off.

    I'm going to have to look in the book some more for that. I appreciate your help!
  13. Jan 29, 2008 #12
    It's the same formula, except that I inserted -9.8, instead of +9.8.
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