Can You Derive Equations Using UAM Principles?

[jensgt]

Homework Statement

Can you come up with (i.e. derive) equations 5-1 and 5-2 on your own using UAM equations? Try it

Homework Equations

This is the equation I need to end up at...

R=Vox/g(Voy+(Voy^+2gh)^1/2

The Attempt at a Solution

I am not really sure where to start. I was trying to use the equations for range and height but it does not seem to be getting me anywhere. This is my first physics course.

 

[tiny-tim]

welcome to pf!

hi jensgt! welcome to pf!

(try using the X² button just above the Reply box )

don't try to memorise (or use) the range or height equations , go back to the standard constant acceleration equations, for the x and y directions (separately)

they'll use the same t …

show us what you get ​

 

[jensgt]

The equations I know are

Vx = Vox + Axt

X = Xo + Voxt + 1/2axt^2

X-Xo = [(Vox + Vx)/2]t

Vx^2 = Vox^2 +2ax(X-Xo)

I am just confused as to where to start. I derived the other one he asked us about this one has me stuck.

 

[tiny-tim]

hi jensgt!

(just got up :zzz:)

I am just confused as to where to start.

use x = x_o + V_xot + 1/2a_xt²

and y = y_o + V_yot + 1/2a_yt²

(obviously, a_x = 0) …

show us what you get ​

(btw, never use "x" for times, it's too confusing … use "*" instead, or nothing at all! )

 

[Nickie]

Hello,

Thank you for your question. I am happy to help you with deriving these equations.

To start, let's define the variables used in the equation you provided. R stands for range, which is the horizontal distance traveled by an object. Vox is the initial velocity in the horizontal direction, g is the acceleration due to gravity, Voy is the initial velocity in the vertical direction, and h is the height of the object.

Now, let's take a look at the equations for range and height. The equation for range is R = Vox * t, where t is the time. The equation for height is h = Voy * t - 1/2 * g * t^2.

To derive the equation R = Vox/g * (Voy + (Voy^2 + 2gh)^1/2), we will need to use these two equations and some algebraic manipulation.

First, let's solve the equation for height for t, which gives us t = (Voy + (Voy^2 + 2gh)^1/2) / g. We can then substitute this value of t into the equation for range, which gives us R = Vox * (Voy + (Voy^2 + 2gh)^1/2) / g.

Next, we can simplify this equation by multiplying both sides by g. This gives us gR = Vox * (Voy + (Voy^2 + 2gh)^1/2).

Finally, we can rearrange this equation to get the desired form of R = Vox/g * (Voy + (Voy^2 + 2gh)^1/2).

Similarly, we can derive the equation R = Vox/g * (Voy - (Voy^2 + 2gh)^1/2) by using the equation for height, h = Voy * t - 1/2 * g * t^2, and following the same steps as above.

I hope this helps. Let me know if you have any further questions or if you need any clarification on the derivation process. Keep up the good work in your physics course!

Best,