(a) With what speed does it leave the roof?
Referring to the image in the link:
Let the positive x-axis go down the incline, with x=0 at the initial location of the piece of ice, and define the y-axis as 90 degrees to the x-axis with positive upwards to the left, with y=0 at the initial location of the piece of ice. Call Θ the angle between the roof and the horizontal ( here, 50 degrees)
The forces on the mass in the x-direction are the component of the weight down the incline (mg sin(Θ)), and the force of friction in the opposite direction (μk × N), where N is the normal force of the incline on the mass. As an equation:
Fx = m × g × sin(Θ) - μk × N
In the y-direction, the forces are the normal force N, and the component of the weight in the opposite direction (mg cos(Θ)):
Fy = N - m × g × cos(Θ)
Fy must equal zero, since there is no acceleration in the y-direction. Therefore:
N = m × g × cos(Θ)
In the x-direction :
m × ax = m × g sin(Θ) - μk × N, so
ax = g × sin(Θ) - μk × g × cos(Θ) = g × (sin(Θ) - μk × cos(Θ))
Plugging in the known values:
ax = 9.8 m/s2 × (sin(50o) - 0.14 × cos(50o)) = 6.625 m/s2
The equation of motion in the x-direction is:
xt = x0 + v0 × t + (1/2) × ax × t2
When the ice leaves the roof at time t1, after sliding a distance D (8m), this simplifies to:
xt1 = D = (1/2) × ax × t12
We can now solve for t1:
t1 = √(2 × D / ax) = 1.554 s
Along the x-axis :
v1(t1) = v0 + ax × t1, where:
v1(t1) is the speed when it leaves the roof, and
v0 = 0; the initial velocity, so
v1 = 0 + 6.625 m/s2 × 1.554 s = 10.30 m/s
Up to and including this point the motion is only in the positive x-direction; there is no velocity in the y-direction.
(b) How far away from the foot of the building does the ice land?
Since we now know the speed and direction of the ice just when it leaves the roof, and we know the vertical distance to the ground, we can determine the distance away from the foot of the building that the ice lands.
To make the calculation simpler, we will switch to using a coordinate system with x0 and y0 both equal to 0 at the point where the ice leaves the roof. We will also have the x-axis parallel to the ground (horizontal in the diagram) with positive values to the left, and the y-axis vertical with positive values upwards.
The equation of projectile motion along the x-axis :
x(t) = x0 + v0,x × t + (1/2) × ax × t2
... where :
v0,x is now the initial velocity, and is the horizontal component of the final velocity in part a) :
v0,x = v1 × cos(Θ) = 6.618 m/s
... and the acceleration in the x-direction (which is now horizontal):
ax = 0
The equation of projectile motion in the y-axis :
y(t) = y0 + v0,y × t + (1/2) × ay × t2
... where :
v0,y is now the initial velocity, and is the vertical component of the final velocity in part a):
v0,y = -v1 × sin(Θ) = -7.887 m/s
... and the acceleration in the y-direction is g:
ay = -g = -9.8 m/s2
Once the ice leaves the roof, the equation of motion for y is:
y(t) = 0 - v1 × sin(θ) × t - (1/2) × g × t2
By solving this quadratic equation for t:
t = { -v1 × sin(Θ) +/- √ [ (v1 × sin(Θ))2 - 2 × g × y(t) } /g
From this equation we can find the value of t for a given value of y(t); using that we will find the final x-value xf, when the ice hits the ground.
We know that the final position of the ice has a y-value yf = -4.0 m at the time tf the ice hits the ground, so by plugging in known values, when can calculate tf:
tf = 0.405 s
In general, in the x-direction:
x(t) = x0 + v0x × t + (1/2) × ax × t2
... so when the ice hits the ground, the x-position is:
xf = v0x × tf + (1/2) × ax × tf2
We know that the final position of the ice has a y-value yf = -4.0 m at the time t=tf when the ice hits the ground; the final x-value xf is what we want to find.
After the ice leaves the roof, the y-equation can then be written:
yt = 0 - v1 × sin(Θ) × t - (1/2) × g × t 2
which can be rewritten:
(1/2) × g × t 2 + v1 × sin(Θ) × t + yt = 0
The solution to this quadratic equation in time is:
t = { -v1 × sin(Θ) +/- √ [ (v1 × sin(Θ)) ^2 - 2 × g × yt ] } /g
tf = 0.405 s ; only the real solution is physically allowed
... so when the ice hits the ground:
xf = x0 + v0,x × tf = 2.68 m
