Is the Faulty Velocity Equation for Projectile Launch Valid?

[DAB]

I have seen this equation used to calculate the velocity of a projectile being launched off a table top.

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.

 

[Dale]

Everything I see has a time component

What are the units of $\sqrt{g/2h}$

 

[DAB]

D and h in meters, g= -9.8 m/sec squared

 

[Dale]

So what are the units of $\sqrt{g/2h}$

 

[DAB]

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).

 

[mfig]

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?

 

[DAB]

The answer (V) is expressed in m/s. meters/second

 

[PeroK]

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?

 

[DAB]

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.

 

[PeroK]

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.

 

[DAB]

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

is equal to

 

[mfig]

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.

 

[Dale]

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.