One side of the roof of a building slopes up at 30.5°. A roofer kicks a round, f

Question

One side of the roof of a building slopes up at 30.5°. A roofer kicks a round, flat rock that has been thrown onto the roof by a neighborhood child. The rock slides straight up the incline with an initial speed of 15.0 m/s. The coefficient of kinetic friction between the rock and the roof is 0.455. The rock slides 10.0 m up the roof to its peak. It crosses the ridge and goes into free fall, following a parabolic trajectory above the far side of the roof, with negligible air resistance. Determine the maximum height the rock reaches above the point where it was kicked in (m).

Explanation / Answer

Given,

theta = 30.5 deg ; u = 15 m/s ; u = 0.455 ; h = 10 m

The net force on the rock at the roof will be:

Fnet = -mg sin(theta) - u mg cos(theta)

ma = -mg (sin(theta) + u cos(theta))

a = -g [sin(theta) + u cos(theta)]

a = -9.81[sin30.5 + 0.455 x cos30.5] = -8.82 m/s^2

from eqn of motion, the velocity while it leaves:

v^2 = u^2 + 2 a s

v = sqrt (u^2 + 2 a s)

v = sqrt (15^2 - 2 x 8.82 x 10) = 6.97 m/s

We know that:

h = vy^2/2g = (6.97 sin30.5)^/2g =

So the maximum height reached will be:

H = 10 sin30.5 + 0.62 = 5.69m

Hence, H = 5.69 m

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