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Homework Help: Horizontal distance travelled at given time

  1. May 26, 2010 #1
    I need to figure out how far an object has travelled at a given time.

    For example purposes the object will be thrown with a velocity of 30 m/s and at an angle of 30 degrees.

    I've figured out the distance/peak height/peak time/time of flight but I don't quite understand how to figure out the horizontal distance travelled at the peak height.

    Distance: 79.5m
    Peak Height: 6.99m
    Peak Time: 1.53s
    Time of Flight: 3.06s

    I can't seem to find the formula with an example. Any help?
  2. jcsd
  3. May 26, 2010 #2


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    Staff: Mentor

    Welcome to the PF.

    Hint: if you neglect air resistance, the horizontal velocity is constant.

    BTW, is this for schoolwork? If so, I can move it to the Homework Help forums for you.
  4. May 26, 2010 #3
    You're going to slap your head when you hear this but since you have neglected air resistance, there are no forces in the horizontal direction. Hint: use sin (theta).
  5. May 26, 2010 #4


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    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi TeamChuck! Welcome to PF! :smile:

    The horizontal speed is constant (because horizontal acceleration is zero), so just use the time of flight that you've found. :wink:
  6. May 26, 2010 #5
  7. May 26, 2010 #6
    Not homework. I'm building a very basic basketball game that requires 3 sets of x,y coordinates: start/peak/end.

    Neglecting air resistance the horizontal distance would be...half the total distance?

    Head slap...more like head meet wall. If this is right. I haven't done any physics since highschool, which was ten years ago.
  8. May 26, 2010 #7


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    Staff: Mentor

    Is this real basketball, or a video game? You may not be able to neglect air resistance, depending on what you are trying to do.
  9. May 26, 2010 #8
    Do you get what to do with sin(theta)?
  10. May 26, 2010 #9
    No, not real. Simple flash game. The user is given a random position on the court and then shoots at the net. The only thing the user controls is the angle and power(velocity). All shots go directly at the basket, its just a matter of whether you under or over throw the basket.

    Heh, or whether you hopefully get it in.
  11. May 26, 2010 #10
    And no, I don't know what sin(theta) is.
  12. May 26, 2010 #11


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    Staff: Mentor

    You wouldn't use sin(theta) anyway....
  13. May 26, 2010 #12
    My bad. I was using vertical as theta = 0. cos(theta)
  14. May 26, 2010 #13
    This question is related to my topic but along a different line.

    What is the equation used to calculate the angle/velocity if you are given the starting x,y position and the x,y coordinates of the peak of the curve?

    Is there even a formula for that? Or would you need to know the time it took to reach the peak as well?
  15. May 27, 2010 #14
    OK, let's back up. You need to start with F=ma where a=v'=x". Also, v=x'. All of these (with the exception of m) are vectors. You will need to understand a little about vectors and basic calculus to solve these equations. If you do not understand any calculus or the concept of a vector, then that is where you need to begin. Try wikipedia on Newton's Laws to get a grasp on these concepts if they are not familiar.
  16. May 27, 2010 #15
    I have a general understanding of Calculus and Newtons three laws.

    Maybe if I restate my question. I've been trying to understand the calculations on my own but I'm having difficulty. I have the x,y(pt.A) starting coordinates and I have the peak x,y(pt.B) coordinates. Because resistence is negligeble, the ending x,y(pt.C) would be twice the distance between pt.A and pt.B.

    I would like to know the initial force and the time that it would take to complete.
  17. May 27, 2010 #16
    What have you calculated? Can you write the equation of motion in terms of the x,y positions, velocities and accelerations?
  18. May 27, 2010 #17
    Starting point would be 0,0(pt.A), peak of the curve would be 50,50(pt.B) and the end point would be 100,0(pt.C). As far as calculations go, I'm not sure. Because I don't have the force/angle I need to work back from the peak/range. I just don't know how to do that.
  19. May 27, 2010 #18
    First, take the angle piece out of it and shoot the ball straight up. What is the force acting on the ball and how fast is it going at apogee?
  20. May 27, 2010 #19
    Well, I wasn't sure if 'apogee' was a typo or not but as best I can figure you mean the farthest distance from the earth? If that's correct, then 50.
  21. May 27, 2010 #20
    What is the vertical speed there? and the force?
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