One side of the roof of a building slopes up at 380 A roofer

One side of the roof of a building slopes up at 38.0°. 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 0.390. 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.

Solution

Let the mass of the rock = m.

Force between rock and roof due to rock’s weight = F = mg.cos= mg.cos 38º

KE = 1/2 mv² = 1/2 * M * (15m/s)² = 112.5m^2/s^2 * M
work done by friction W = µmgcos*d = 0.390 * M * 9.8m/s² * cos38º * 10m = M*30.11m²/s²
increase in PE = mgh = M * 9.8m/s² * 10m * sin38º = M *60.33m²/s²
Then the rock has KE = M(112.5 - 30.11- 60.33)m²/s² = 22.06m²/s² * M = ½Mv²
so v =6.64 m/s
We have the launch angle and velocity:
max. height = (V·sin)² / (2g) = (6.64m/s * sin38)² / 19.6m/s² = 0.852m
plus the height it gained along the roof = 10m * sin38 = 6.156 m
yields a total rise h = 6.156+ 0.852m = 7.008m

maximum height the rock reaches = 7.008 m