Answer:
a) 5 m/s downwards
b) 17.86 m/s
c) 24.82 m/s
d) 0.228
Explanation:
We can set the frame of reference with the origin on the top of the building and the X axis pointing down.
The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
V(t) = V0 + a * t
In this case
X0 = 0
V0 = -5 m/s
a = 9.81 m/s^2
To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.
0 = -5 * t + 1/2 * 9.81 * t^2
0 = t * (-5 + 4.9 * t)
One of the solutions is t = 0, but this is when the rock was thrown.
0 = -5 + 4.69 * t
4.9 * t = 5
t = 5 / 4.9
t = 1.02 s
Replacing this in the speed equation:
V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)
When the rock is at 15 m above the street it is 15 m under the top of the building.
15 = -5 * t + 1/2 * 9.81 * t^2
4.9 * t^s -5 * t - 15 = 0
Solving electronically:
t = 2.33 s
At that time the speed will be:
V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s
When the rock is about to reach the ground it is at 30 m under the top of the building:
30 = -5 * t + 1/2 * 9.81 * t^2
4.9 * t^s -5 * t - 30 = 0
Solving electronically:
t = 3.04 s
At this time it has a speed of:
V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s
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Power is work done per unit of time.
The work in this case is:
L = Ff * d
With Ff being the friction force, this is related to weight
Ff = μ * m * g
μ: is the coefficient of friction
L = μ * m * g * d
P = L/Δt
P = (μ * m * g * d)/Δt
Rearranging:
μ = (P * Δt) / (m * g * d)
1 horsepower is 746 W
20 minutes is 1200 s
μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228
north of east