# Trajectory of a horizontally fired projectile

1. Oct 3, 2012

### szimmy

1. The problem statement, all variables and given/known data
For a lab that we did I need to draw the path of a horizontally fired projectile using a graph. When I look at the equation I derived it doesn't look like it will draw the path correctly.

3. The attempt at a solution
Δx = Vox*t
T = Δx / Vox
-h = Voy*t + .5*a*t2
-h = Voy*(Δx / Vox) + .5*a* (Δx / Vox)2
-h = 0 * (Δx / Vox) + .5*-g*(Δx2 / Vox2)
-h = -g/2 * (Δx2 / Vox2)
h = (g*Δx2)/ (2*Vox2)
h = (g)/ (2*Vox2) * Δx2
y = (g)/ (2*Vox2) * x2
y = c*x2
c = (g)/ (2*Vox2)

h being the height the projectile was fired from above the ground. This looks almost right, but shouldn't I have a negative somewhere in my constant? The shape it should draw is the right half of an upside down parabola, should I not have used -h even though the projectile is moving under the starting point? Any help would be greatly appreciated, I'm a bit puzzled at this point.

2. Oct 3, 2012

### PhanthomJay

The equation is y = y_o + V_oy(t) + 1/2at^2. Where y_o is the initial position of the projectile , h, at time t = 0 and position x = 0. You forgot the y_o term , and then you threw in an extra minus sign in front of the y. The portion on the right side of the equation determines the signage of y. y and h are not the same.