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!

Frisbee physics problem

  1. Oct 10, 2007 #1
    One side of the roof of a building slopes up at 39.0°. A student throws a Frisbee onto the roof. It strikes with a speed of 15.0 m/s and does not bounce, but slides straight up the incline. The coefficient of kinetic friction between the plastic and the roof is 0.380. The Frisbee slides 10.0 m up the roof to its peak, where it goes into free-fall, following a parabolic trajectory with negligible air resistance. Determine the maximum height the Frisbee reaches above the point where it struck the roof.


    I think this one is the hardest question I have seen yet. In this question it says
    the frisbee slides 10.0m up the roof to its peek, does it mean frisbee goes up diagonally
    not vertically. If it goes diagonally I got the wrong answer for it. So how I can solve
    this problem? Also, if a frisbee strikes the roof, does a frisbee stays 15.0m/s or it was
    instantaneously slows down after it hits the roof?
     
  2. jcsd
  3. Oct 10, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. It slides up the roof, so it moves along the incline of the roof.
    Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.
    I would assume that the frisbee loses no speed when it hits the roof--its speed when it begins sliding up the roof is 15 m/s.
     
  4. Oct 10, 2007 #3
    Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.

    I would assume that the frisbee loses no speed when it hits the roof--its speed when it begins sliding up the roof is 15 m/s.[/QUOTE]



    Here are the calculations for this question

    [x-component for F] = 0 - [Kinetic friction of force] - [mg*sin39] = ma (1)
    [y-component for F] = [Normal force] - [mg*cos39] = 0 (2)


    [Kinetic friction of force] = [Coefficient of kinetic friction]*[Normal force] (3)


    then find "normal force" of (2)

    [Normal force] = [mg*cos39] (4)


    then substitute (4) into (3)

    [Kinetic friction of force] = [Coefficient of kinetic friction]*[mg*cos39] (5)


    then substitute (5) into (1)

    -[Coefficient of kinetic friction]*[mg*cos39] - [mg*sin39] = ma


    then find acceleration

    a = [-([Coefficient of kinetic friction]*[mg*cos39])-(mg*sin39)] / m

    = (-[Coefficient of kinetic friction]*[g*cos39])-(g*sin39)

    = -g([(Coefficient of kinetic friction)*cos39]+[sin39])

    = -9.0707m/s^2



    Now I have to find time

    v = at + [Initial velovity] *Now do the anti-derivative
    d = [(at^2)/2] + [(Initial velocity)*t] + [Initial displacement]
    d = [(at^2)/2] + [(Initial velocity)*t] *Initial displacement is gone because its zero


    then substitute a into this equation which becomes

    10 = [(-4.5353m/s^2)*(t^2)] + [15.0m/s*t]
    0 = [(-4.5353m/s^2)*(t^2)] + [15.0m/s*t] - 10


    then use quadratic equation

    (-b[+-]sqrt[(b^2)-(4ac)])/(2a)
    (-15[+-]sqrt[(15^2)-4(-4.5353)(-10)])/(2*[-4.5353])

    t = 0.9258s, 2.3815s I choose 0.9258s for this calculation


    substitute t into the velocity equation

    v = at + [Initial velovity]
    = (-9.0707m/s^2)*(0.9258s) + 15.0m/s
    = 6.6020m/s


    then substitute v into this formula

    [Final velocity] = [Initial velocity] + at
    0 = [Initial velocity]*sin39 - gt


    find t

    t = ([Initial velocity]*sin39)/g


    then subsitute t into this formula

    d = ([Initial velocity]*t) + (0.5*a*t^2)
    h = (Initial velocity*sin39)*[([Initial velocity]*sin39)/g] - [(0.5*g)(([Initial velocity]*sin39)/g)]^2
    h = [([Initial velocity]^2)*((sin 39)^2)]/(2g)
    h = [((6.6020m/s)^2)*((sin39)^2)]/(2*9.81m/s^2)
    h = 0.8798m


    then find a height of the roof

    h = (sin 39)*10
    h = 6.2932m

    then add it together

    [Maximum height] = 0.8798m + 6.2932m
    [Maximum height] = 7.1730m = 7.17m
     
    Last edited: Oct 10, 2007
  5. Oct 10, 2007 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Don't forget the x-component of gravity.
     
  6. Oct 10, 2007 #5
    What do you mean x-component of gravity?

    I don't understand what you mean?
     
    Last edited: Oct 10, 2007
  7. Oct 10, 2007 #6
    So it becomes like this:


    [x-component for F] = 0 - [Kinetic friction of force] - [mg*sin39] = ma (1)
    [y-component for F] = [Normal force] - [mg*cos39] = 0 (2)
     
  8. Oct 10, 2007 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Yes. Looks good.
     
  9. Oct 10, 2007 #8
    So I recalculate the question (look above) do I do all the calculations right?
     
  10. Oct 10, 2007 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, looks OK.

    But you could save yourself some effort if you learned another kinematic formula:

    [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Frisbee physics problem
  1. A frisbee (Replies: 1)

  2. Physics problems (Replies: 4)

  3. Physic Problem (Replies: 3)

  4. Physic Problem. (Replies: 3)

  5. A physics problem. (Replies: 5)

Loading...