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Interesting Projectile Motion problem

  1. Oct 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A person throws a ball off a roof that is 8 meters high with an angle of 45 degrees from the positive x axis. The ball is supposed to land in the center of a square that is centered 9 meters away from the base of the roof. With which initial velocity must the person throw the ball so that it lands from it is supposed to?





    2. Relevant equations

    Vi = unknown
    Vx = 9/Ttot
    Vx = Vi cos (45)
    Vy = Vi sin (45)
    h = 9 meters
    y = distance from the initial height of the ball to the height that the ball peaks.
    Ttot = the total time the ball is in there air.

    3. The attempt at a solution

    Ttot = t1 + t2 where t1 = the time it takes the ball to peak and t2 = the time it takes the ball to fall from that maximum height.

    t1 = -Vy/g
    t2 = sqrt ( (2h + 2y)/g)

    t3 = -Vy/g + sqrt ( (2h + 2y)/g) = 9/Vx

    When replaced with everything I'm given, I get the following equation. Keep in mind we're trying to solve for Vi.

    t3 = -Vi sin(45)/g + sqrt ( (2(8) + 2(-Vi sin (45)/2g)/g) = 9/(Vi cos (45) )

    From here, I'm not sure how I can solve for Vi. Is this problem impossible to solve with what I'm given or am I just making it a lot harder than it needs to be? I'd appreciate any help ya'll can give me!
     
  2. jcsd
  3. Oct 3, 2009 #2
    I'd just use the component method.

    This is what you end up with

    Vx = Vocos([tex]\Theta[/tex])
    Vy = Vosin([tex]\Theta[/tex])

    Because the angle is 45o, what can you assume about the X and Y components of the velocity?

    From that you can use the kinematics formulas to mess around with the stuff you have and find other information.

    For instance:

    d=Vot+(1/2)at2

    You know what the final vertical displacement will be, and from this you can substitute and solve.

    Or you could try using other kinematics equations.

    Just remember that X only has one equation: Dx=Vxt
     
  4. Oct 3, 2009 #3
    No you don't know what the final vertical displacement will be because you don't know the initial velocity.

    Yes I know the velocity components of Vi are the same.

    Did you look at the problem?
     
  5. Oct 3, 2009 #4
    You said it needs to land in a square at ground level from a 8 METER high roof. So your displacement is -8.
     
  6. Oct 3, 2009 #5
    No because if it was initial velocity UPWARD, the initial height will be greater than 8 meters. The additional height will be equal to Vy^2/2g.

    Since I don't know Vy without knowing Vi, I don't have the complete vertical displacement.
     
  7. Oct 3, 2009 #6

    lewando

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    Gold Member

    I agree with Mattowander's initial assertion that he is making this harder than it needs to be. Setting up the x-y origin at the edge of the roof (implied by Lancelot), concentrate on solving the following equations at the specific time T when the projectile is at (X, Y) = (9, -8)

    X = vxi * T [x component of motion]

    Y = (vyi * T) -9.8 T^2 [y component of motion]

    vxi = vyi
     
  8. Oct 3, 2009 #7

    lewando

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    Gold Member

    Sorry, forgot the 1/2
    Y = (vyi * T) -(9.8 T^2)/2 [y component of motion]
     
  9. Oct 3, 2009 #8
    http://en.wikipedia.org/wiki/Displacement_%28vector%29" [Broken] is the distance from A to B, not the distance you travelled in getting there.
     
    Last edited by a moderator: May 4, 2017
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