Thread: Frisbee physics problem View Single Post
P: 85
 Quote by Doc Al 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.
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

(-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