Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Velocity of projectile launcher

  1. Mar 18, 2007 #1
    I recently did an expirement for my physics class in which we used a projectile launcher to fire a ball across a distance, where it would hit a piece of carbon paper.

    I've been asked to solve for the initial velocity, (Vi), and the only known information is:
    Vertical Acceleration = -9.81 m/s^2
    Horizontal Displacement = 3.885m
    Delta Y (Height Ball was launched above the ground) = 1.015m
    Theta = 9 degrees

    I'm not asking anyone to solve this for me, but I'm really stuck. I was wondering if anyone could tell me the steps I must do, or even better, an equation that I could use to solve this.

    Any help is greatly appreciated.

    Thank you.
  2. jcsd
  3. Mar 18, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

  4. Mar 18, 2007 #3
    It might help to understand the problem if you were to do the problem backwards first... Assume some sort of initial velocity, then do the calculations to find out where the ball would land. Look at how you can work that problem backwards to find the initial velocity; then go ahead and do so for the data you have.
  5. Mar 18, 2007 #4
    I'd rather not do a 'trial and error' approach, as I do have to show real work.
  6. Mar 18, 2007 #5
    so, you have a parabola for the projectile, correct?

    notice that

    make substitution and get rid of the t.

    now, you have two points and you know the last coefficient of a parabola, and you know theta. what is the other point on the parabola that you know x and y?

    you'll have one equation, one variable, just solve it.
    Last edited: Mar 18, 2007
  7. Mar 18, 2007 #6


    User Avatar
    Homework Helper

    Noone suggested such an approach. The link from post #2 should pretty much solve your problem.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook