# Frisbee physics problem

1. Oct 10, 2007

### shiri

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. Oct 10, 2007

### 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.

3. Oct 10, 2007

### shiri

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
4. Oct 10, 2007

### Staff: Mentor

Don't forget the x-component of gravity.

5. Oct 10, 2007

### shiri

What do you mean x-component of gravity?

I don't understand what you mean?

Last edited: Oct 10, 2007
6. Oct 10, 2007

### shiri

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)

7. Oct 10, 2007

### Staff: Mentor

Yes. Looks good.

8. Oct 10, 2007

### shiri

So I recalculate the question (look above) do I do all the calculations right?

9. Oct 10, 2007

### Staff: Mentor

Yes, looks OK.

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

$$v^2 = v_0^2 + 2 a \Delta x$$