A man starts from rest at the top of a 10 meter tall slide (10 meters off of the

Question

A man starts from rest at the top of a 10 meter tall slide (10 meters off of the ground). The slide goes down at a 45 degree angle, approaches ground level, and then curves up so that the slide makes a 40 degree angle with the ground, ending 2.5 meters above the ground, pointing upward. How far will the man travel when he flies off the slide at a 40 degree angle, from 2.5 meters above flat ground? (Where should we put a pool for him to land in, relative to the end of the slide?) Assume a coefficient of friction between man and waterslide to be 0.1. Assume that the ground is perfectly level. Knowing the mass of the man is 80kg. (You will have to approximate the energy lost from friction.) A man starts from rest at the top of a 10 meter tall slide (10 meters off of the ground). The slide goes down at a 45 degree angle, approaches ground level, and then curves up so that the slide makes a 40 degree angle with the ground, ending 2.5 meters above the ground, pointing upward. How far will the man travel when he flies off the slide at a 40 degree angle, from 2.5 meters above flat ground? (Where should we put a pool for him to land in, relative to the end of the slide?) Assume a coefficient of friction between man and waterslide to be 0.1. Assume that the ground is perfectly level. Knowing the mass of the man is 80kg. (You will have to approximate the energy lost from friction.) A man starts from rest at the top of a 10 meter tall slide (10 meters off of the ground). The slide goes down at a 45 degree angle, approaches ground level, and then curves up so that the slide makes a 40 degree angle with the ground, ending 2.5 meters above the ground, pointing upward. How far will the man travel when he flies off the slide at a 40 degree angle, from 2.5 meters above flat ground? (Where should we put a pool for him to land in, relative to the end of the slide?) Assume a coefficient of friction between man and waterslide to be 0.1. Assume that the ground is perfectly level. Knowing the mass of the man is 80kg. (You will have to approximate the energy lost from friction.)

Explanation / Answer

length of the 1st slide = L1 = 10/sin45

frictional force on 1st slide = u*m*g*cos45

work done by friction on 1st slide = Wf1 = u*m*g*cos45*10/sin45

Wf1 =u*m*g*10/tan45

length of the 2nd slide = L2 = 2.5/sin40

frictional force on 2nd slide = u*m*g*cos40

work done by friction on 2nd slide = Wf2 =

Wf2 = u*m*g*4.5/tan40

from work energy theorem

m*g*h1 - Wf1 - wf2 = m*g*h2 + 0.5*m*v^2

h1 = 10 m

h2 = 2.5 m

(9.8*10)-((0.1*9.8*10)/tan45) - ((0.1*9.8*2.5)/tan40) = (9.8*2.5)+(0.5*v^2)

v= 11.02 m

horizantal component of velocity after leaving the 2nd slide = vx = v*cos40 = 8.44 m/s

maximum horizantal distance R = v^2*sin(2*theta)/g

R = (11.02^2*sin80)/(9.8)

R = 12.20 m    <<<------answer

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