Trajectory of a horizontally fired projectile

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SUMMARY

The discussion focuses on deriving the trajectory of a horizontally fired projectile using the equations of motion. The participant initially derived the equation h = (g)/(2*Vox²) * Δx², which represents the height of the projectile as a function of horizontal distance. However, confusion arose regarding the correct application of signs in the equation, particularly concerning the initial height (y_o) and the negative sign associated with downward motion. The correct equation for the projectile's path is y = y_o + V_oy(t) + 1/2at², emphasizing the importance of correctly identifying the initial position and the direction of motion.

PREREQUISITES
  • Understanding of kinematic equations for projectile motion
  • Familiarity with the concepts of initial velocity (V_ox and V_oy)
  • Knowledge of gravitational acceleration (g) and its effects on motion
  • Ability to manipulate algebraic equations and understand parabolic functions
NEXT STEPS
  • Study the derivation of projectile motion equations in physics textbooks
  • Learn about the significance of initial conditions in kinematic equations
  • Explore graphing techniques for parabolic trajectories in physics simulations
  • Investigate the effects of varying initial heights on projectile motion
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of projectile motion and graphing parabolic trajectories.

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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