Is the Faulty Velocity Equation for Projectile Launch Valid?

AI Thread Summary
The discussion centers on the validity of a simplified velocity equation for projectiles launched from a height, specifically D = horizontal distance, h = height, and g = gravitational acceleration. Participants clarify that the equation, while seemingly simple, incorporates time implicitly through the term √(g/2h), which represents the time taken for an object to fall a distance h. This means that the equation does indeed have a time component, contrary to initial assumptions. The conversation emphasizes the importance of understanding the units involved, confirming that the expression yields velocity in meters per second. Overall, the equation is valid, but it is essential to recognize its underlying relationship with time.
DAB
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I have seen this equation used to calculate the velocity of a projectile being launched off a table top.
upload_2018-2-24_11-51-5.png
Where D=horizontal distance travelled, h=table height, and g=gravitational acceleration. Nowhere else can I find a calculation of velocity based on only two distances and GA. Everything I see has a time component or one of the velocity variables (V-i, V-f). Is this equation valid? It seems too simple. I don't care about exploring air resistance or drag coefficients, just a simple velocity at lauch calculation.
 

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DAB said:
Everything I see has a time component
What are the units of ##\sqrt{g/2h}##
 
D and h in meters, g= -9.8 m/sec squared
 
So what are the units of ##\sqrt{g/2h}##
 
I guess I'm not understanding your question. g is the GA constant of -9.8 meter/second squared and h is the height of the table. Do you want a measurement for h? One measurement I have for this is h= -1.25 meters and D= 2.4 meters. That's all the information that seems to be required to calculate V (velocity at launch).
 
He is asking for the units of the expression. Here is a hint:

##\sqrt\frac{\Big(\frac{m}{s^2}\Big)}{m}##

When cancellations are done, what units do you have left?
 
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The answer (V) is expressed in m/s. meters/second
 
DAB said:
I have seen this equation used to calculate the velocity of a projectile being launched off a table top. View attachment 220987 Where D=horizontal distance travelled, h=table height, and g=gravitational acceleration. Nowhere else can I find a calculation of velocity based on only two distances and GA. Everything I see has a time component or one of the velocity variables (V-i, V-f). Is this equation valid? It seems too simple. I don't care about exploring air resistance or drag coefficients, just a simple velocity at lauch calculation.
Can you check whether the equation is true in an example case?
 
How would I know for sure? I'd have to be able to measure the time and the time measurement would be so small that I'd not be able to accurately measure it. I guess I could quesstimate it, but it would be under a second, so pretty hard to measure accurately.
 
  • #10
DAB said:
How would I know for sure? I'd have to be able to measure the time and the time measurement would be so small that I'd not be able to accurately measure it. I guess I could quesstimate it, but it would be under a second, so pretty hard to measure accurately.

I meant try some ##v## and ##h##. Calculate ##D## and check the formula.

I assume your algebra isn't good enough to prove or disprove the general case. The next best thing is to try a few examples.
 
  • #11
This was a simple question. I just wanted to know if anyone had ever seen this equation before. It seems too simple to calculate velocity especially when other equations have one calculating both horizontal and vertical components to arrive at an answer.
Maybe you can answer one last question for me. Tell me if my algebra is correct in calculating that
upload_2018-2-24_11-51-5-png.png
is equal to
upload_2018-2-24_15-24-38.png
 

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  • #12
What people were hinting to earlier is that the units of the expression with the radical are ##\frac{1}{seconds}##. So that is the time aspect you were wondering about.

Specifically, the expression

##\sqrt\frac{2h}{g}##

is the time it takes an object to fall a distance h. So if D is the distance traveled horizontally when the object first touches the floor and this is divided by the travel time, we should get an expression for the horizontal velocity when the object left the table.
 
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  • #13
DAB said:
I guess I'm not understanding your question.
Apparently not. The quantity ##\sqrt{g/2h}## has units of ##1/s## in SI units. So in your OP when you said “Everything I see has a time component” it turns out that ##\sqrt{g/2h}## is the time component you are used to seeing.
 
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