Question

A mischievous SUOC student has climbed on the roof of a snow-covered building and is trying to hit her friend with snowballs as he walks through the Quad. She throws them at an angle of 20◦ above the horizontal at a speed of v0 = 5 m/s. The building has a height h = 6 m. He decides to throw a snowball back at her. He’s standing a distance d from the side of the building, and throws a snowball at an angle θ above the horizontal at a speed v0. However, the snowball slips out of his hand when he throws it, and it doesn’t go very fast – instead of hitting her on top of the building, it hits the side of the building.

1. Draw a cartoon of the problem, making clear your coordinate system and origin, and labelling interesting things.

2. Write expressions for x(t), y(t), vx(t), and vy(t), substituting in variables that you know.

3. Write a sentence in terms of your algebraic variables that will let you figure out how far above the ground the snowball hits the side of the building.

4. Based on your sentence, figure out how far above the ground the snowball hits the building. Your answer should be in terms of v0, θ, d, and g.

5. He doesn’t give up, though, and throws another snowball at her – again at an angle θ above the horizontal. He throws this one harder, and it hits her feet as she stands on the edge of the building. Write a sentence in terms of your algebraic variables that will let you figure out how fast he had to throw it.

6. Now, based on your previous sentence, figure out the initial speed of the second snowball he threw.

Answer 1

Answer 2

javascriptCopy code                                 θ
                                /
                               /
                             / h
                            /|
       o         x        / |
       |               /  | d
       |              /   |
       |             /    |      /|            /     |
  v0 / |           /      |    /  |          /       |   /   |         /        |  /____|_______ /_________|   Building    Ground

In the cartoon, the origin (0, 0) is at the base of the building. The x-axis is horizontal, and the y-axis is vertical, with positive y in the upward direction. The snowball is thrown at an angle θ above the horizontal, and the student is standing a distance d from the side of the building.

2. Expressions for x(t), y(t), vx(t), and vy(t):

The equations of motion for the snowball in the x and y directions are:

• x(t) = v0 * cos(θ) * t

• y(t) = h + v0 * sin(θ) * t - (1/2) * g * t^2

• vx(t) = v0 * cos(θ)

• vy(t) = v0 * sin(θ) - g * t

Where:

• v0 is the initial speed of the snowball (same for x and y directions)

• θ is the launch angle above the horizontal

• h is the height of the building

• g is the acceleration due to gravity (approximately 9.81 m/s^2)

3. Sentence to find how far above the ground the snowball hits the side of the building:

To find how far above the ground the snowball hits the side of the building, we need to determine the value of y when x = d. In other words, we need to find the y-coordinate of the snowball when it reaches the horizontal distance d from the building's side.

4. Calculation to find how far above the ground the snowball hits the building:

Substitute x = d into the expression for y(t):

y(d) = h + v0 * sin(θ) * t - (1/2) * g * t^2

Since x = d and vx(t) = v0 * cos(θ), we can rearrange the equation for x(t) to solve for t:

t = d / (v0 * cos(θ))

Now, substitute this value of t into the equation for y(d):

y(d) = h + v0 * sin(θ) * (d / (v0 * cos(θ))) - (1/2) * g * (d / (v0 * cos(θ)))^2

Simplify the expression:

y(d) = h + d * tan(θ) - (g * d^2) / (2 * v0^2 * cos^2(θ))

So, the height above the ground where the snowball hits the side of the building is given by h + d * tan(θ) - (g * d^2) / (2 * v0^2 * cos^2(θ)).

5. Sentence to find how fast he had to throw the second snowball:

To find how fast he had to throw the second snowball, we need to determine the value of v0 (initial speed) for the given launch angle θ, such that the snowball hits her feet when she stands on the edge of the building. In other words, we need to solve for v0 when y(t) = h.

6. Calculation to find the initial speed of the second snowball:

Set y(t) = h and solve for v0:

h + v0 * sin(θ) * t - (1/2) * g * t^2 = h

Since we are solving for v0, we need to express t in terms of v0 and θ. Use the expression for vy(t) = v0 * sin(θ) - g * t to solve for t:

t = (v0 * sin(θ)) / g

Now, substitute this value of t into the equation for y(t):

h + v0 * sin(θ) * ((v0 * sin(θ)) / g) - (1/2) * g * ((v0 * sin(θ)) / g)^2 = h

Simplify the expression:

v0^2 * sin^2(θ) / g - (1/2) * v0^2 * sin^2(θ) / g = 0

Combine like terms:

(1/2) * v0^2 * sin^2(θ) / g = 0

Now, solve for v0:

v0^2 = 0

v0 = 0 (Since v0 cannot be negative)

Therefore, the initial speed (v0) of the second snowball that hits her feet on the edge of the building is 0 m/s. This means that the snowball did not have any initial speed and was likely dropped from the edge of the building.

Answer 3

1. Cartoon of the problem:

yamlCopy code                     Snowball's path
                        /|
                       / | h
                      /  |
                     /   | 
                    /    |
                   /     | 
                  /      |
                 /       |                / θ     | 
  Snowball       -------|-------
  Thrown at       d
  angle θ

Coordinate system:

• Let the origin (0, 0) be at the base of the building where the person is standing.

• The x-axis is horizontally along the ground, with the positive direction pointing towards the right.

• The y-axis is vertically upwards, with the positive direction pointing upwards.

1. Expressions for x(t), y(t), vx(t), and vy(t):

• x(t) = d + v0 * cos(θ) * t (horizontal distance = initial distance + horizontal velocity * time)

• y(t) = h + v0 * sin(θ) * t - (1/2) * g * t^2 (vertical height = initial height + vertical velocity * time - (1/2) * acceleration due to gravity * time^2)

• vx(t) = v0 * cos(θ) (constant horizontal velocity)

• vy(t) = v0 * sin(θ) - g * t (vertical velocity = initial vertical velocity - acceleration due to gravity * time)

1. Sentence to figure out how far above the ground the snowball hits the side of the building: The snowball hits the side of the building when y(t) = 0.

2. Calculate how far above the ground the snowball hits the building: Set y(t) = 0 and solve for t: 0 = h + v0 * sin(θ) * t - (1/2) * g * t^2 Solve for t: (1/2) * g * t^2 - v0 * sin(θ) * t - h = 0

Use the quadratic formula: t = (-(-v0 * sin(θ)) ± √(v0 * sin(θ))^2 - 4 * (1/2) * g * (-h)) / 2 * (1/2) * g t = (v0 * sin(θ) ± √(v0 * sin(θ))^2 + 2 * g * h) / g

We choose the positive root (time can't be negative): t = (v0 * sin(θ) + √(v0 * sin(θ))^2 + 2 * g * h) / g

Now, substitute this value of t into y(t) to find the height above the ground when the snowball hits the side of the building: y(t) = h + v0 * sin(θ) * ((v0 * sin(θ) + √(v0 * sin(θ))^2 + 2 * g * h) / g) - (1/2) * g * ((v0 * sin(θ) + √(v0 * sin(θ))^2 + 2 * g * h) / g)^2

1. Sentence to figure out how fast he had to throw the second snowball: The speed of the second snowball when it hits her feet is vy(t) = 0.

2. Calculate the initial speed of the second snowball he threw: Set vy(t) = 0 and solve for v0: 0 = v0 * sin(θ) - g * t

Solve for v0: v0 = g * t / sin(θ)

Now, substitute the value of t (calculated from part 4) into the above equation to find the initial speed v0 of the second snowball.