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Homework Help: Initial Speed and Projectile Motion

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data

    So it's a projectile motion problem. I draw a graph showing the parabolic trajectory of the ball, with the start point at the origin. It's final x = 24.8m and final y = 0m

    2. Relevant equations

    We have three constant acceleration equations we use in my course:

    [tex]\vec{v}_{fx} = \vec{v}_{ix} + \vec{a}_{x}t[/tex]

    Ensuring everything is in the same direction (vertical or horizontal), final velocity = initial velocity + acceleration * time

    [tex]\vec{x}_{f} = \vec{x}_{i} + \vec{v}_{ix}t + \frac{1}{2}\vec{a}_{x}t^2[/tex]

    Looks like an integral. Final position = initial position + initial velocity * time + half acceleration * t squared

    [tex]\vec{v}_{fx}^2 = \vec{v}_{ix}^2 + 2\vec{a}_{x}Δ\vec{x}[/tex]

    Final velocity squared = initial velocity squared + double acceleration * displacement

    3. The attempt at a solution

    My prof posted a video meant to give hints that will help us get started on the problem. I can try linking it here, though I'm not sure it'll work:


    Basically what it says is that, the motion is parabolic, I know the max height, and we know from class that the velocity at a projectile at max height is 0.

    I then draw a graph with only half the parabola, starting at (x,y) = (0,4.98) and ending at (24.8, 0).

    Then I have:

    [tex]\vec{a}_{y} = -9.8m/s/s[/tex]
    [tex]\vec{v}_{iy} = 0m/s[/tex]
    [tex]\vec{y}_{i} = 4.98m[/tex]
    [tex]\vec{x}_{f} = 24.8m[/tex]
    [tex]t = ?[/tex]
    [tex]vf = ?[/tex]

    I use the second equation I listed above to find t:

    [tex]0m = 4.98m + (0m/s)t + 1/2(-9.8m/s/s)t^2[/tex]
    [tex]0m = 4.98m - (4.98m/s/s)t^2[/tex]
    [tex](4.9m/s/s)t^2 = 4.98m[/tex]
    [tex]t = 1.0081302s[/tex]

    I keep in mind that this is only the time for half the parabola, so if I use this for the whole thing, I will need to double it.

    We know from class also that the velocity at the end of a projectile's path is equal to but opposite in direction of the initial velocity. We also know that if we find the x and y components of the final velocity, we can add them to obtain the final velocity.

    For the y component, I use equation 2 from above:

    [tex]\vec{v}_{fy} = \vec{v}_{iy} + \vec{a}_{y}t[/tex]
    [tex]\vec{v}_{fy} = 0m/s + (-9.8m/s/s)(1.0081302s)[/tex]
    [tex]\vec{v}_{fy} = -9.87968m/s[/tex]

    For the x component, I use equation 2 (acceleration along x is always 0m/s/s for projectiles, we learned in class):

    [tex]\vec{x}_{f} = \vec{x}_{i} + \vec{v}_{ix}t + \frac{1}{2}\vec{a}_{x}t^2
    [tex]24.8m = 0m + (\vec{v}_{ix})(1.0081302s) + 0[/tex]
    [tex]24.8m = (1.0081302s)\vec{v}_{ix}[/tex]
    [tex]\vec{v}_{ix} = 24.599997m/s[/tex]

    This is where my main problem is. I'm guessing I did all of the above correctly, because it was simply subbing into the equations. I'm just not sure how to get the initial SPEED when I have the VELOCITY components.

    Hopefully this is easy enough to understand, and thanks for all who read.
    Last edited: Sep 22, 2013
  2. jcsd
  3. Sep 22, 2013 #2
  4. Sep 22, 2013 #3

    Doc Al

    User Avatar

    Staff: Mentor

    That first point is not part of the parabola. It reaches max height at half the horizontal range.



    Don't forget that the time is for half the motion.

    If you have the components of any vector, use the Pythagorean theorem to find the magnitude.
  5. Sep 22, 2013 #4
    Thanks to both of you. I don't know how I managed to miss that while reviewing my notes.

    I also seem to have done the x component wrong. The initial x position was 12.4m, not 0m, which gave me the correct answer with the pythagorean theorem. Thanks again.

    Now if I want the initial velocity of the ball, I can just add both velocity components, correct?
  6. Sep 22, 2013 #5

    Doc Al

    User Avatar

    Staff: Mentor

    As long as you "add" them as vectors, sure.
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