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.
What velocity does the milk exit the nipple with to be able to have that trajectory?
From ChatGPT:
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.