Trajectory of a horizontally fired projectile

AI Thread Summary
The discussion focuses on deriving the path of a horizontally fired projectile, with the user attempting to graph the trajectory using their derived equation. They express confusion over the signs in their calculations, particularly regarding the height (h) and its relationship to the initial position (y_o). The correct equation should account for y_o and clarify the negative sign associated with downward motion. The key takeaway is that y and h represent different concepts, and proper attention to the signs is essential for accurately depicting the projectile's path. Understanding these distinctions will help ensure the graph reflects the expected upside-down parabolic shape.
szimmy
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Homework Statement


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.

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.
 
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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.
 
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