1. An arrow is shot from the top of a 50 m castle wall at 18 m/s horizontal to the

ground. How far from the base of the castle wall will it land?
2. A cannon with a muzzle velocity of 45m/s is shot a various angles to get
maximum distance. The general knows that if it is shot straight up it will land on
top of him, but at an angle too small the ball will hit the ground too soon. What is
the best angle to have the cannonball travel the greatest horizontal distance. To
solve this try 15°, 30°, 45°, 60° and 75°.
3. A hockey puck of mass 0.115 kg moving at 35m/s strikes a rubber octopus thrown
on the ice by an over exuberant fan. The octopus has a mass of 0.265 kg. The
puck and the octopus slide along the ice together. Assuming no friction, find the
velocity of the pair.

1) y: t = sqrt(2*50/9.8)

x = 18t
2) Or just know that it is 45 because R = v^2 sin2theta/g
3) As much as I hate the Red Wings:
m1v1 = (m1+m2)vf

1. To solve this problem, we need to consider the horizontal and vertical motions separately. First, let's focus on the horizontal motion. We know that the initial horizontal velocity of the arrow is 18 m/s. Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the motion.

Next, let's consider the vertical motion. We can use the kinematic equation for vertical motion:

y = y0 + v0yt - (1/2)gt^2

where:
y = vertical position (final position - height from the base of the wall)
y0 = initial vertical position (0 since the arrow is shot from the top of the wall)
v0y = initial vertical velocity (0 since there is no initial vertical velocity)
g = acceleration due to gravity (-9.8 m/s^2)
t = time

Since we want to find the horizontal distance, we need to find the time it takes for the arrow to hit the ground. We can use the equation:

y = 0 = y0 + v0yt - (1/2)gt^2

Simplifying the equation, we get:

- (1/2)gt^2 = -y0
(1/2)gt^2 = y0

Solving for t, we have:

t = sqrt(2y0) / sqrt(g)

Substituting the values, we have:

t = sqrt(2 * 50 m) / sqrt(9.8 m/s^2)
t = sqrt(100) / sqrt(9.8)
t = 10 s / 3.13
t ≈ 3.19 s

Now that we know the time, we can find the horizontal distance using:

x = v0x * t

Substituting the values, we have:

x = 18 m/s * 3.19 s
x ≈ 57.42 m

Therefore, the arrow will land approximately 57.42 m from the base of the castle wall.

2. To determine the best angle for the cannonball to travel the greatest horizontal distance, we need to analyze the projectile motion. The horizontal and vertical motions are independent, so we can analyze them separately.

Let's calculate the horizontal distance for each angle using the following equation:

x = (v^2/g) * sin(2θ)

where:
x = horizontal distance
v = initial velocity (45 m/s)
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of projection

Let's calculate the horizontal distances for each angle:

For θ = 15°:
x = (45^2/9.8) * sin(2 * 15°)
x ≈ 207.56 m

For θ = 30°:
x = (45^2/9.8) * sin(2 * 30°)
x ≈ 412.35 m

For θ = 45°:
x = (45^2/9.8) * sin(2 * 45°)
x ≈ 630.58 m

For θ = 60°:
x = (45^2/9.8) * sin(2 * 60°)
x ≈ 829.30 m

For θ = 75°:
x = (45^2/9.8) * sin(2 * 75°)
x ≈ 895.04 m

From the calculations, we can see that the cannonball will travel the greatest horizontal distance when it is launched at an angle of approximately 75°.

3. To find the resulting velocity of the puck-octopus pair, we need to apply the principle of conservation of momentum. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it.

Let's assume that the initial velocity of the puck is v1 and the initial velocity of the octopus is v2. Since we're assuming no friction, the final velocity of the pair would be the same for both, denoted by v.

Using the conservation of momentum equation:

m1v1 + m2v2 = (m1 + m2)v

where:
m1 = mass of the puck (0.115 kg)
m2 = mass of the octopus (0.265 kg)
v1 = initial velocity of the puck (35 m/s)
v2 = initial velocity of the octopus (0 m/s)
v = final velocity of the pair

Substituting the values, we have:

(0.115 kg)(35 m/s) + (0.265 kg)(0 m/s) = (0.115 kg + 0.265 kg)v

Simplifying the equation, we get:

4.0255 kg m/s = 0.38 kg v

Solving for v, we have:

v ≈ 10.607 m/s

Therefore, the velocity of the puck-octopus pair is approximately 10.607 m/s.