I am hoping one of my fellow Ags can assist me with how to solve this physic problem. My son is taking high school physics and was given the following question, but we are not able to find out how to solve it and the class has no book nor is the teacher really given out any notes. Any help would be appreciated.

Suppose you adjust your garden hose nozzle for a fast stream of water you point the nose vertically up at a height of 1.8 meters above the ground because why not? When you quickly turn off the nozzle you hear the water striking the ground next to you for another 2.6 seconds. What is the water speed as it leaves the nozzle?

My Ag '25 has the answer but I can't post a photo of his work (not subscriber). How can I share it w you?

Coming from my '27 student and I have no freaking clue but says answer is Vo = 5.94 m/s.

I'll try typing it…but it's tricky to get the notation correct
Formula is …..
Change in y = v_o(t) + 1/2 a(t^2)
-1.8m = v_o(2.6s) + 1/2(-9.81m/s^2)(2.6s)^2

it is incorrect to assume that the final velocity is zero for obvious reasons, since the water is splashing on the ground

solving the equation I mentioned gives v_o = 12.06 m/s

Sure wish I could post a photo of his work.

Text me at 4 six 9 9 six 4 5 two 1 two and I'll send it.
BTHO high school physics AND miami

Oh this is good if he's wrong... I can give him **** about something I don't know jack about. Haha

LoL!

Per ChatGPT

To find the speed at which the water leaves the nozzle, we can use the equations of motion. When the water is shot vertically upward, it reaches a maximum height before falling back down. The total time it takes to return to the ground can be calculated by:
t_total = 2 * t_upward,
where t_upward is the time it takes for the water to reach its maximum height. To find t_upward, we can use the following kinematic equation:
h = (1/2) * g * t_upward^2,
where:

• h is the height above the ground (1.8 meters),
• g is the acceleration due to gravity (approximately 9.81 m/s).

Let's solve for t_upward first:
1.8 = (1/2) * 9.81 * t_upward^2.
Now, solve for t_upward:
t_upward^2 = (1.8 * 2) / 9.81,
t_upward^2 = 0.366,
t_upward 0.366,
t_upward 0.605 seconds.
Now, we know it takes approximately 0.605 seconds for the water to reach its maximum height. To find the speed at which it leaves the nozzle, we can use the following equation of motion for vertical motion:
v = g * t_upward.
Plug in the values:
v = 9.81 m/s * 0.605 s,
v 5.94 m/s.
So, the water leaves the nozzle with a speed of approximately 5.94 meters per second

I'm the '25 kid mentioned above.

GPT screwed up here, it thought that 1.8m is the max height of the hose, but it's actually the starting position.

Answer provided by Aggielax00 (my mom ) is right. Gig 'em.

aggielax00 said:

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I'll try typing it…but it's tricky to get the notation correct
Formula is …..
Change in y = v_o(t) + 1/2 a(t^2)
-1.8m = v_o(2.6s) + 1/2(-9.81m/s^2)(2.6s)^2

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This is correct. The displacement is 1.8 meters downward from the time it leaves the hose to the time it hits the ground.