# Max horizontal displacement

1. Oct 5, 2005

### TickleMeElma

You buy a plastic dart gun, and you wnt to find its max. horizontal range. you shoot the gun straight up, and it takes 4s for the dart to land back at the barrel. What is the maximum horizontal range of your gun?

I got the initial velocity, which is 19.6 m/s. I know I have to incorporate vectors. The initial velocity in the horizontal direction is 19.6m/s (cos x) and in the vertical direction it is 19.6m/s (sin x). Assuming x is the magnitude of the angle. I know that the total displacement in the vertical direction is 0.

Thanks so very much.

2. Oct 5, 2005

### hotvette

Hint: What value for the angle x will produce the maximum range?

Last edited: Oct 5, 2005
3. Oct 5, 2005

### whozum

Your $\vec{v_i}$ vector points at whatever direction you fire at. To fire for maximum distance you want to fire at a 45 degree angle to the horizontal, which means you'll have velocity vector [itex] \vec{v_i} = 19.6m/s @ 45 \ deg [/tex]. Find how far that goes.

4. Oct 6, 2005

### TickleMeElma

Thank you so much for your help - yeah, somehow, I missed the sample problem that dealt with the apparently widely-known concept of 45 degrees being the angle that yields the max horizontal replacement. Went back and reworked it with your suggestions and those of the book and got ~39m. :)

5. Oct 6, 2005

### hotvette

It can be sort of intuitive. If the angle is 90 (i.e. straight up), the range is zero. If the angle is zero, the range is zero. Thus, by reasoning, the max would be 1/2 way between the two, or 45 degrees.

Actually, there is a rigorous way to prove it using the equations of motions for a projectile y(t) and x(t). By setting y = 0 and solving for t you get the time it takes to complete the motion. Using this t in the equation for x, you get what's called the range equation, which gives you x as a function of angle. Since you want max x, differentiate x with respect to theta, set it equal to zero, and solve for theta.