Thanks! It was fun to write at the time. I was doing a double major of History of Art and Philosophy but the HoA was definitely inflating my grade average!
Cano did a good job of convincing the church he was being sincere! Probably would have been excommunicated or executed if they suspected him of herasy.
Edit: he personally described his work as being "edifying" lol
I can't remember if I stumbled across it at a video store or caught it on cable or something, but even though I was a KitH fan I had no idea they were making a movie. Absolutely love it though and even with familiar elements and some returning characters it didn't feel like just a rehash of the show. I was bummed to learn (from the 2022 "Comedy Punks" documentary) that they were totally miserable making it and there was bad blood between Dave and everyone else.
Me: Water exits a nozzle at coordinate (0, 1.7m) and hits the ground at (2m, 0). Assume the water exits the nozzle at 90 degree (i.e. straight to the right).
How can we determine the speed of the water?
(here I made some rough assumptions based on the image, and assumed that her feet were at (0, 0)
Answer:
The speed of the water exiting the nozzle can be determined using the principles of projectile motion.
Determine the time of flight:
The water is projected horizontally from a height of 1.7 meters and hits the ground at 2 meters horizontally. The vertical motion is governed by the equation:
[
y = y_0 + v_{y0} t - \frac{1}{2} g t^2
]
Here, ( y_0 = 1.7 ) meters, ( v_{y0} = 0 ) m/s (since there is no initial vertical velocity), and ( g = 9.8 ) m/s² (acceleration due to gravity).
The water hits the ground when ( y = 0 ):
[
0 = 1.7 - \frac{1}{2} \times 9.8 \times t^2
]
Solving for ( t ):
[
1.7 = 4.9 t^2
]
[
t^2 = \frac{1.7}{4.9}
]
[
t = \sqrt{\frac{1.7}{4.9}} \approx 0.59 \text{ seconds}
]
Determine the horizontal velocity:
The horizontal distance covered is 2 meters. Since horizontal velocity ( v_x ) is constant (there is no horizontal acceleration), the distance can be given by:
[
x = v_x t
]
Here, ( x = 2 ) meters and ( t \approx 0.59 ) seconds:
[
v_x = \frac{2}{0.59} \approx 3.39 \text{ m/s}
]
Therefore, the speed of the water exiting the nozzle is approximately 3.39 m/s.