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Projectile and force! !

  1. Jan 12, 2010 #1
    projectile and force! :D !

    1. The problem statement, all variables and given/known data

    A 3.00 kg projectile must be launched across a room so that it lands 8.50 meters from the launching point. If the ceiling height is 2.90 meters can this be done? If so, what force must be applied over a time period of 0.111 seconds and at what angle for the projectile to land 8.50 meters from the launching point?

    2. Relevant equations

    3. The attempt at a solution
    I don't think it's possible.. :(
  2. jcsd
  3. Jan 13, 2010 #2
    Re: projectile and force! :D !

    sure it is/imagine that the projectile is a cannonball!

    I don't know that there is a unique solution unless we are to assume it nearly grazes the ceiling.
  4. Jan 13, 2010 #3
    Re: projectile and force! :D !

    I think it's as long as it doesn't surpass 2.9m.
  5. Jan 13, 2010 #4
    Re: projectile and force! :D !

    That's the thing, there are any number of ways to hit the target, because there is no limit on the force we can apply. Imagine tossing a wad of paper into a wastebasket a few feet away--you could throw it on a straight line, almost a straight line, or arc it in at various angles. It doesnt say we need to graze the ceiling hence the number of ways to hit the target are infinite. So take your pick, realizing that gravity will change the y velocity and the horizontal velocity dictates the flight time.
  6. Jan 13, 2010 #5
    Re: projectile and force! :D !

    To solve it I need to find vi for the y direction. I don't know how to derive the equation: x=vi*t+.5*a(t)^2 to solve for vi.
  7. Jan 13, 2010 #6
    Re: projectile and force! :D !

    Getting ready to retire, but lets see if I can help.

    When you say derive--I think you mean solve. Google quadratic equation.

    But before we go there, lets start by assuming that we need to graze the ceiling.

    That means the peak height is 2.9 meters.

    The question then is how fast must an object be traveling initially in the Y direction to hit that peak?

    The quickest way is to use the eqn Vf^2-Vi^2=2*a*y where a is gravitational acceleration and Vf = 0 (it has peaked and is traveling at zero speed before beginning the descent). Can you get Vy from that?
  8. Jan 13, 2010 #7
    Re: projectile and force! :D !

    I got 7.54 ms. I don't know how to apply force with an angle and time.
  9. Jan 13, 2010 #8
    Re: projectile and force! :D !

    Gonna fast forward here as I may not be online much longer.

    I assume you were able to compute how fast the upward velocity was to peak at 2.9m.

    How long did it take to get there?

    One thing that will help you a lot in solving problems of this type is to understand that the time to travel up is the same as it is down.

    So lets imagine the projectile is simply dropped from 2.9m. How long does it take to travel 2.9m?

    The eqn you have posted simplifies in this instance to y=1/2at^2. So 2*y/a=t^2

    Again a is the acceleration due to gravity (9.8m/s^2).

    OK? Now visualize the flight of the projectile--there are two phases--one up and one down. The time spent in each is the same.

    So the total time of flight is twice what you computed above.

    That time of flight is the same as that needed to traverse the horizontal distance. From this compute a horizontal velocity--this doesn't change as there are no forces acting in this direction.

    pretty much done: we have an initial y velocity, an initial (and steady) x velocity.

    Only one thing left: we need to accelerate the projectile in a time of 0.11 sec to reach the initial combined x-y velocity. This will be v below.

    v=at. f=ma
  10. Jan 13, 2010 #9
    Re: projectile and force! :D !

    I get 7.6m/s Don't forget the dash! ms is not m/s. Hope the other post helps.
  11. Jan 13, 2010 #10
    Re: projectile and force! :D !

    How do I find the angle then? Do I make another triangle with vx and xy as the legs? I ended with F=357N at 53.7 degrees, is this correct?
  12. Jan 13, 2010 #11
    Re: projectile and force! :D !

    The force is going to be applied in line with the initial angle which can be found as you suggest:

    tan (theta)= Vy/Vx

    Don't have time to check your work, but you can always test it by trying it, ie

    Find a. Multiply by 0.11 sec to get Vinitial. Then decompose into to Vy and Vx. Vy you calculated above, etc.
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