Calculating Maximum Height of a Sliding Frisbee on a Sloped Roof

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In summary: No, just substitute the values into the original equation.In summary, a Frisbee slides 10.0m up the roof to its peak, where it goes into free-fall, following a parabolic trajectory with negligible air resistance.
  • #1
shiri
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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?
 
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  • #2
shiri said:
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.
Yes. It slides up the roof, so it moves along the incline of the roof.
If it goes diagonally I got the wrong answer for it. So how I can solve
this problem?
Find the net force on the frisbee as it slides up. Then you can find the acceleration and its speed as it flies off.
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?
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
Doc Al said:
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] = mathen 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 zerothen 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] - 10then 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 calculationsubstitute t into the velocity equation

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

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

t = ([Initial velocity]*sin39)/gthen 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 = 0.8798mthen 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:
  • #4
shiri said:
Here are the calculations for this question

[x-component for F] = 0 - [Kinetic friction of force] = ma (1)
[y-component for F] = [Normal force] - [mg*cos39] = 0 (2)
Don't forget the x-component of gravity.
 
  • #5
Doc Al said:
Don't forget the x-component of gravity.

What do you mean x-component of gravity?

I don't understand what you mean?
 
Last edited:
  • #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)
 
  • #7
Yes. Looks good.
 
  • #8
Doc Al said:
Yes. Looks good.

So I recalculate the question (look above) do I do all the calculations right?
 
  • #9
shiri said:
So I recalculate the question (look above) do I do all the calculations right?
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]
 

What is the physics behind throwing a Frisbee?

The physics behind throwing a Frisbee involves principles of aerodynamics, specifically the Bernoulli's principle and the Magnus effect. When the Frisbee is thrown, it creates a lift force due to the difference in air pressure above and below the Frisbee's curved surface. This lift force allows the Frisbee to stay in the air and have a curved flight path due to the Magnus effect, which is caused by the Frisbee's spin.

What factors affect the flight of a Frisbee?

The flight of a Frisbee is affected by various factors such as the angle of release, the velocity of the throw, the spin of the Frisbee, the weight and shape of the Frisbee, and the air resistance. These factors can impact the distance, speed, and trajectory of the Frisbee's flight.

Why does a Frisbee sometimes curve during its flight?

A Frisbee curves during its flight due to the Magnus effect, which is caused by the Frisbee's spin. As the Frisbee spins, it creates an area of low pressure on one side and high pressure on the other side. This pressure difference creates a lift force that causes the Frisbee to curve in the direction of its spin.

How can the principles of physics be applied to improve Frisbee throwing?

Understanding the physics behind Frisbee throwing can help improve throwing techniques. For example, by adjusting the angle of release and the spin of the Frisbee, one can control the direction and curve of its flight. Additionally, knowing about air resistance can help in choosing the right Frisbee for different weather conditions.

What other real-world applications can be found for the physics of Frisbee throwing?

The principles of physics used in Frisbee throwing can be applied to other sports such as disc golf and ultimate frisbee. Additionally, the understanding of aerodynamics and lift forces can be applied in the design of aircrafts and other flying objects.

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