Part C

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a pitch of 21 degrees. If the edge of the roof is 11.0 above ground, what is the speed of the snow when it hits the ground?

Well, snow projectiles from slippery roofs sounds like an icy version of a cannonball competition! Let's calculate the speed of that snow chunk's daring descent, shall we?

First, we need to break down the problem into two components: one in the horizontal direction and the other in the vertical direction.

In the horizontal direction, the speed remains constant because there's no force acting on the snow chunk. So, the horizontal component of the velocity will stay the same throughout.

In the vertical direction, the snow chunk will accelerate downwards due to gravity. The acceleration due to gravity, let's call it "g", is approximately 9.8 m/s².

Now, let's dig into the calculations. Since the pitch angle of the roof is given as 21 degrees, we can use trigonometric functions to find the vertical component of the velocity.

The vertical component of the velocity (v_y) can be found using the formula: v_y = v * sin(theta), where v is the total velocity and theta is the angle of inclination.

Since we only need to know the vertical component of the velocity, we can use the following equation to solve for v_y:

v_y = v * sin(21 degrees)

Now, the initial vertical velocity (v_y) would be zero since the snow chunk starts from rest. So, we can use the following formula to find the time it takes for the snow chunk to reach the ground:

v_y = g * t

Substituting v_y = 0 and g = 9.8 m/s², we get:

0 = 9.8 * t

Solving for t, we find that it takes the snow chunk 0 seconds to reach the ground. Well, that's impressive! It's a snow chunk with teleportation abilities.

Now, let's move on to the horizontal component of the velocity:

v_x = v * cos(21 degrees)

We know that there's no horizontal acceleration, and since the snow chunk travels horizontally with a constant velocity, the horizontal distance traveled (d_x) is given by:

d_x = v_x * t

Plugging in v_x = v * cos(21 degrees) and t = 0, we find that the horizontal distance traveled by the snow chunk is also zero. It seems like our snow chunk prefers to stay perched on the roof rather than venture out into the world.

In summary, based on the given information, it appears that the snow chunk will neither gain horizontal nor vertical velocity and will magically stay on the roof forever. Guess it wants to become one with the house!

To find the speed of the snow when it hits the ground, we can use the concept of conservation of energy.

First, let's calculate the potential energy of the snow when it is at the ridge of the roof:

Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)

Given:
Height of the roof (h) = 11.0 meters
Gravitational acceleration (g) = 9.8 m/s^2 (approximately)

Now, the mass of the chunk of snow is not provided in the question, but we don't need it specifically to determine the speed of the snow when it hits the ground. This is because the mass will cancel out when calculating the speed using the conservation of energy equation.

Next, let's calculate the potential energy at the ridge of the roof:

PE = m * g * h

Substituting the given values:
PE = m * 9.8 * 11.0

Now, let's consider the conservation of energy:
Potential energy (at the ridge of the roof) = Kinetic energy (when it hits the ground)

Since the snow starts from rest, its initial kinetic energy is zero.

So, potential energy = kinetic energy

PE = (1/2) * m * v^2 (where v is the speed of the snow when it hits the ground)

Setting the two equations equal to each other and solving for v:

m * 9.8 * 11.0 = (1/2) * m * v^2

Simplifying the equation:

9.8 * 11.0 = (1/2) * v^2

v^2 = (2 * 9.8 * 11.0) / 1

v^2 = 215.6

Taking the square root of both sides:

v = √215.6

v ≈ 14.7 m/s

Therefore, the speed of the snow when it hits the ground is approximately 14.7 meters per second.