I want to find Max horizontal range without

[Saladsamurai]

using [tex]\frac{v_0^2}{g_0}=R_{max}[/tex] for this problem:

You buy a toy dart gun and you want to calculate its Max Horizontal Range. You fire the gun straight upward and find that it takes 2.9 seconds for the dart to leave the barrel and then return to the barrel.

Can this be done without the formula. I already tried this:

[tex]V_y=v_{oy}+a_yt[/tex] to find that [tex]v_{oy}=14.21[/tex] I used v_f=0 and t=1.45 to find v_y initial. Now that should be equal to the magnitude of just plain [tex]v_0[/tex] right? or is that a false assumption?

 

[learningphysics]

Yes, so the vo for the gun is 14.21m/s. That's correct.

Why can't you use that formula?

Strange... because I don't see how to calculate the maximum range without ultimately using or deriving that equation...

 

[Saladsamurai]

Yes, so the vo for the gun is 14.21m/s. That's correct.

Why can't you use that formula?

Strange... because I don't see how to calculate the maximum range without ultimately using or deriving that equation...

Well, I can...but I don't want to; I want to derive it but I am having trouble arriving at the same number I get by using the range formula.

I am assuming that I get max range at 45 degrees...is that not correct?

So range would equal [tex](v_0)_x*t=14.21\cos45*t[/tex] Right? BUt what do I use for t? I must have to eliminate it..huh?

 

[Saladsamurai]

I got it. I used [tex]v_y=v_{oy}+at[/tex] to find t and then multiplied it by 2 to get [tex]\frac{14.21sin45}{4.9}=t[/tex] in [tex]\Delta x=v_{ox}*t[/tex] and got the exact same number...sweet.

Thanks again LP,
Casey