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Projectile Motion Lab vertical/horizontal motion?

  1. Oct 21, 2012 #1
    http://www.foothillfalcons.org/ourpages/auto/2008/10/19/62799693/Rolling%20Ball%20Lab.pdf [Broken]

    I am doing a lab in class where we set up a ramp on a table and measured the distance the ball travels on the horizontal part of the track. We then rolled the a marble down the ramp and did trials on the time it took to travel the horizontal distance (catching the ball before it fell off the table)

    I used the formula V= D/T to get speed of the ball as it rolled across the table.

    How do I use kinematics equations to determine the horizontal and vertical distance from the table at which the ball will hit the floor.?

    Vertical:

    Vi =? d = ?
    Vf = ? a = ?
    t = ?
    What equation should I use and how do I get the variables?

    Horizontal:
    Vi = 0 d = ?
    Vf = ? a = -9.8 m/s^2
    t= ?

    If anyone could help me even a little I would appreciate it!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 21, 2012 #2
    What to you think the values for the variables are in each dimension? The values you have for the horizontal (X) dimension part are incorrect. You only need one equation to solve for one variable vertically and only need one equation to solve for one variable horizontally.
    Have you studied projectiles/two dimension problems yet or is this designed to introduce the concept?
     
  4. Oct 21, 2012 #3
    We just started this lesson. I' not sure what to do
     
  5. Oct 21, 2012 #4
    Focus just on the horizontal dimension first. As the marble leaves the table, what is the initial velocity? What is its horizontal velocity just before it hits the floor (Vf)?
     
  6. Oct 23, 2012 #5
    Okay, so firstly you need to calculate the velocity at which the ball is moving down the ramp.

    Use v2 = u2 + 2as

    Since we're talking about a falling object, 'a' turns to 'g'

    After you find the velocity, you want to find the time it's going to travel from the edge of the table to the floor.

    s = vit + (1/2)at2

    In this case, 's' is the distance from the top of the table to the ground.

    Finally, the horizontal distance it will travel is the easy part.

    x = vt

    And done.
     
    Last edited: Oct 23, 2012
  7. Oct 23, 2012 #6
    This is a ramp, not a free-fall situation where motion is parallel to the force of gravity. There is some trig to do to calculate 'a'.
     
  8. Oct 23, 2012 #7
    Right, if we set up right triangle for the ramp, 'g' would be the short side. You need to use the angle at which the ramp is elevated and use sin(theta) = g/h to find 'h' as a vector of the ramp.
     
  9. Oct 23, 2012 #8
    You don't need to do all of that. You already know how fast the ball is moving when it leaves the table from step 4 of the procedure. The ball will continue to move at this velocity in the horizontal dimension until it hits the floor. So for the horizontal dimension, there is no acceleration once the ball leaves the ramp. Show us what you think the rest of the variables equal.
     
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