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Projectile being launched off a table at an angle?

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    We did an experiment where we launched a ball off a table at an initial velocity of 1.80 m/s, at an angle of 10 degrees. We measured the distance in the x-axis the ball travelled 0.743 m. Using this information, we are supposed to calculate a predicted value for the distance in the y-axis (initial height). Since we did the experiment, we have an experimental value of 1.12 m for the height, but my teacher wants us to calculate a predicted value for initial height and compare using absolute % error ((predicted - experimental) / experimental).

    2. Relevant equations

    vix = (vi)cos(10)
    vfy=(vi)sin(10)
    ax = 0
    ay= -9.81
    t = ?
    dx= 0.743 m
    dy = ?
    3. The attempt at a solution

    I calculated vix to be 1.773 m/s and viy to be 0.3125667 m/s. I'm unsure how to calculate time, and also am unsure if there is a vi for y-axis (my teacher said I did the question wrong because I used dy= vit + 1/2at^2 and cancelled out vit because I believed vi was 0 but the ball is being launched so there is a vi for y).
     
  2. jcsd
  3. Feb 18, 2014 #2
    The equation you used wrongly is OK, just use it correctly. Since ##v_{iy}## is not zero, as you said, do not set it to zero.
     
  4. Feb 18, 2014 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    hi megmar95! welcome to pf! :smile:
    the initial height is zero

    do you mean maximum height? :confused:

    (and is the table flat? and is the ball landing on the table again, or on the floor?)
    yes, vi is at an angle, so viy = visin10°
     
  5. Feb 18, 2014 #4
    Ya I meant the distance travelled by the ball in the y-axis altogether (range)
    The ball was launched off the end of a ramp angled at 10 degrees placed on top of a flat table at the very edge of it, and landed on the floor.
    I tried using the (vi)sin(10) and plugged it into the equation d=vit+1/2at^2, so d= (0.3125)(0.419) + 1/2(9.81)(0.419)^2 and I got an answer of 0.992 m for predicted height. Afterwards I used the % error equation, and got a percent error of 11.4%, does that sound correct?
     
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