Answer:
A) The maximum height above the roof reached by the rock can be found using the formula:
h = (v₀²sin²θ)/(2g)
where v₀ is the initial velocity (32.0 m/s), θ is the angle of the initial velocity (29.0°), and g is the acceleration due to gravity (9.81 m/s²).
Plugging in the values, we get:
h = (32.0²sin²29.0)/(2(9.81)) = 31.1 m
Therefore, the maximum height above the roof reached by the rock is 31.1 meters.
B) The vertical component of the velocity just before the rock strikes the ground is:
vᵥ = v₀sinθ - gt
where t is the time it takes for the rock to reach the ground.
We can find t by using the formula:
h = v₀sinθt - (1/2)gt²
where h is the height of the building (14.0 m). Rearranging this formula and solving for t, we get:
t = (v₀sinθ + sqrt((v₀sinθ)² + 2gh))/g
Plugging in the values, we get:
t = (32.0sin29.0 + sqrt((32.0sin29.0)² + 2(9.81)(14.0)))/9.81 = 4.01 s
Therefore, the vertical component of the velocity just before the rock strikes the ground is:
vᵥ = 32.0sin29.0 - 9.81(4.01) = -14.3 m/s
Note that the negative sign indicates that the velocity is directed downwards.
C) The horizontal distance from the base of the building to the point where the rock strikes the ground can be found using the formula:
d = v₀cosθt
Plugging in the values, we get:
d = 32.0cos29.0(4.01) = 96.4 m
Therefore, the horizontal distance from the base of the building to the point where the rock strikes the ground is 96.4 meters.