1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Projectile Motion

  1. Oct 6, 2005 #1
    I did an experiment which involved a few trials. Here are the results:

    distance rolled on table: 100 cm (all three trials) .
    height of table: 86.4 cm
    time on table: 1.1 s (trial 1), 1.2 s (trial 2), 1.3 s (trial 3) .
    horizontal distance traveled in air: 36.1 cm (trial 1), 34.8 cm (trial 2), 34.8 cm (trial 3) .

    Ok so (1) What is the velocity the ball rolled with on the table? So [tex] v = \frac{d}{t} [/tex]. I got 91 cm/s (trial 1), 83 cm/s (trial 2), 83 cm/s (trial 3).

    (2) The vertical component of the balls velocity when it leaves the edge of the table would be 0. Is this correct?

    (3) Calculate the time that it takes the ball to reach the floor once it has left the table. Ok so I would use [tex] d = v_{y}_{0}t + \frac{1}{2}gt^{2} [/tex]. [tex] v_{y}_{0} = 0, d = 86.4 cm [/tex]. Are these the correct values for the variables?

    (4) Find the horizontal distance that the ball should travel from the time it left the table until it hit the floor for the three trials. So would I use [tex] x = x_{0} +v_{x}_{0}t + \frac{1}{2}a_{x}t^{2} [/tex]? So would I use the 91 cm/s, 83 m/s, and 83 m/s, for the velocities, and 0.42 seconds for the time?

    Thanks
     
    Last edited: Oct 6, 2005
  2. jcsd
  3. Oct 6, 2005 #2
    You are mixing up units for part (3), make sure you convert to one set of units.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Projectile Motion
Loading...