
#1
Oct1007, 02:25 PM

P: 85

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 freefall, 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
Oct1007, 02:44 PM

Mentor
P: 40,872





#3
Oct1007, 04:14 PM

P: 85

I would assume that the frisbee loses no speed when it hits the roofits speed when it begins sliding up the roof is 15 m/s.[/QUOTE] Here are the calculations for this question [xcomponent for F] = 0  [Kinetic friction of force]  [mg*sin39] = ma (1) [ycomponent 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 antiderivative 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 



#5
Oct1007, 04:22 PM

P: 85

I don't understand what you mean? 



#6
Oct1007, 05:00 PM

P: 85

So it becomes like this:
[xcomponent for F] = 0  [Kinetic friction of force]  [mg*sin39] = ma (1) [ycomponent for F] = [Normal force]  [mg*cos39] = 0 (2) 



#8
Oct1007, 05:46 PM

P: 85




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