I want to find Max horizontal range without

AI Thread Summary
To calculate the maximum horizontal range of a toy dart gun, the initial velocity (v_0) was determined to be 14.21 m/s after analyzing the dart's flight time. The discussion highlights the challenge of deriving the maximum range without using the standard formula, which relates initial velocity and gravitational acceleration. It was confirmed that the maximum range occurs at a launch angle of 45 degrees, leading to the equation for range as (v_0)_x * t. The time (t) was calculated using vertical motion equations, confirming the derived range matched the expected results. Ultimately, the conversation emphasizes the relationship between projectile motion principles and the derivation of range calculations.
Saladsamurai
Messages
3,009
Reaction score
7
using \frac{v_0^2}{g_0}=R_{max} 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:

V_y=v_{oy}+a_yt to find that v_{oy}=14.21 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 v_0 right? or is that a false assumption?
 
Physics news on Phys.org
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...
 
learningphysics said:
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 (v_0)_x*t=14.21\cos45*t Right? BUt what do I use for t? I must have to eliminate it..huh?
 
Last edited:
I got it. I used v_y=v_{oy}+at to find t and then multiplied it by 2 to get \frac{14.21sin45}{4.9}=t in \Delta x=v_{ox}*t and got the exact same number...sweet.

Thanks again LP,
Casey
 
Last edited:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top