EK Physics 1001 #586

[mtravis2190]

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Hey everyone! I've been working through the Examkracker's Physics 1001 question series and #586 doesn't look quite right to me. There is an image of an open container with a spigot about two-thirds of the way down. It shows water draining out from it and the height of the water spewing out is 1/4h with a horizontal distance of d. The question asks If the container is filled with water, and h is 20 m, the distance d is: a.) 5m, b.) 10m, c.) 20m, or d.) 40m.
There is an accompanying image. I hope it is clear enough.
The answer is C.

I hope someone can help! I didn't quite like the answer key on the question. Let me know if anyone wants to know what the answer key explanation said; it was a bit long to type, so I'll hold off on it unless anyone asks for it.

Thank you!

 

[brood910]

Answer is right.

sqrt 2gh = velocity of the water leaving, where h is the distance from the water surface to the hole.

So, V = 20.

Then, find the flight time by using 5 = 5t^2 (1/4h = 5 = change in height)
T= 1.
Range = vt = 20 x 1 = 20.

 

[mtravis2190]

Answer is right.

sqrt 2gh = velocity of the water leaving, where h is the distance from the water surface to the hole.

So, V = 20.

Then, find the flight time by using 5 = 5t^2 (1/4h = 5 = change in height)
T= 1.
Range = vt = 20 x 1 = 20.

Hi there! Thank you for the reply. I think what I'm confused by is the fact that the equation sqrt 2gh = velocity gives us velocity in the vertical direction. Doesn't it (since the equation applies acceleration of gravity)? So how can we use that velocity to find range?

 

[brood910]

No, it gives you the velocity of the water when it is leaving the hole.
That means, it's horizontal.

Think of a car jumping off a cliff without any vertical velocity. It's the same situation.

 

[mtravis2190]

No, it gives you the velocity of the water when it is leaving the hole.
That means, it's horizontal.

Think of a car jumping off a cliff without any vertical velocity. It's the same situation.

Alright! Thank you! I really appreciate the prompt responses! 🙂

 

[mtravis2190]

Hello! I'm back! I've been trying to wrap my head around what you said about something falling off with just horizontal velocity and no vertical velocity to justify using the sqrt 2gh = velocity formula. However, from several questions that I did from the EK physics 1001 book on projectile motion and from introductory physics experience, I can't seem to make sense of using this formula to find horizontal velocity since there is no acceleration in the x-direction. I recall using this formula often in cases such as finding max height of a projectile after calculating for velocity in the y-direction from a given initial velocity of a projectile.

I'd appreciate if you could clarify this doubt of mine. Sorry for bugging you (and the others following the thread) again!

 

[brood910]

Before using any "shorten" equations, you should know where they are derived from.
sqrt 2gh = v comes from v2 = v2 + 2ax for projectile, when initial velocity is 0.
For water leaving the hole, it is NOT from the same equation..
It comes from pgh = 1/2pv2 (Bernoulli's equation).

Remember, for physics, DO NOT just memorize equations. Understand where they come from.

 

[mtravis2190]

Before using any "shorten" equations, you should know where they are derived from.
sqrt 2gh = v comes from v2 = v2 + 2ax for projectile, when initial velocity is 0.
For water leaving the hole, it is NOT from the same equation..
It comes from pgh = 1/2pv2 (Bernoulli's equation).

Remember, for physics, DO NOT just memorize equations. Understand where they come from.

Oh ok! That makes better sense now. But I still have another question. Sorry! How could 'g' still apply to horizontal motion?

 

[brood910]

Oh ok! That makes better sense now. But I still have another question. Sorry! How could 'g' still apply to horizontal motion?

You keep getting confused by this because you are trying to incorporate kinematic equations into pressure equations.

 

[Cmdr_Shepard]

Oh ok! That makes better sense now. But I still have another question. Sorry! How could 'g' still apply to horizontal motion?

I had some free time at work so I did the question for you and wrote some directions in there (photo attached).

 

[mtravis2190]

I had some free time at work so I did the question for you and wrote some directions in there (photo attached).
View attachment 178213

Wow! Thank you so much for this. Seeing this definitely confirmed the numbers that I crunched and formulae I used, however, while I worked through the question and even after seeing your solution, I'm still unsure about a few "why's". For instance, why is vertical velocity zero when we are looking to calculate time? Is it because that's the max height point, where PE is max and KE is at its minimum? Secondly, I see how working through Bernoulli you end up with sqrt 2gh = velocity, but I still feel uncomfortable with using g (vertical acceleration) for finding horizontal velocity. But I guess I should just accept it seeing as Bernoulli proves it.
Thank you for spending time in solving the question so thoroughly for me! 🙂