A tin can is filled with water to a depth of 37 cm. A hole 11 cm above the bottom of the can produces a stream of water that is directed at an angle of 32 degrees above the horizontal.

Find the range of this stream of water. x=?
Find the maximum height of this stream of water. ymax=?

To find the range (x) of the stream of water, we need to determine how far it travels horizontally before hitting the ground.

First, let's consider the vertical motion of the water stream. We can use the equation of motion:
y = y0 + v0y * t - (1/2) * g * t^2

where:
- y is the vertical position of the water stream
- y0 is the initial vertical position (depth of the water in the can)
- v0y is the initial vertical velocity (vertical component of the water stream velocity)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time

In this case, y0 = 37 cm = 0.37 m, v0y is the vertical component of the velocity (v0) times the sin of the launch angle (θ), and g = 9.8 m/s^2.

Let's calculate the initial vertical velocity (v0y):
v0y = v0 * sin(θ) = v * sin(θ)

Now, let's calculate the time it takes for the water stream to reach the maximum height (t_max). At the maximum height, the vertical velocity is zero. Using the equation:
v_y = v0y - g * t

When the vertical velocity is zero, v_y = 0:
0 = v0y - g * t_max

From this equation, we can solve for t_max:
t_max = v0y / g

Once we have the time it takes to reach the maximum height, we can calculate the maximum height (y_max):
y_max = y0 + v0y * t_max - (1/2) * g * t_max^2

To find the horizontal range (x), we need to find the time it takes for the water stream to hit the ground. This is done by calculating the total time of flight (t_total). The time of flight can be found by solving the equation:
y = y0 + v0y * t - (1/2) * g * t^2

For calculating the horizontal range, y would be zero, since it hits the ground. So we have:
0 = y0 + v0y * t_total - (1/2) * g * t_total^2

Using the quadratic formula, we can solve for t_total:
t_total = [v0y + sqrt(v0y^2 + 2 * g * y0)] / g or t_total = [v0y - sqrt(v0y^2 + 2 * g * y0)] / g

Finally, we can calculate the range (x) using the equation:
x = v0x * t_total

However, we do not have the value of v0x (horizontal component of velocity) yet. But we can find it using the equation:
v0x = v0 * cos(θ)

Now, we have all the information needed to solve for x (range) and ymax (maximum height).