Projectile motion equation help

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SUMMARY

The discussion focuses on deriving the vertical displacement equation (Dy) as a function of horizontal displacement (Dx) using the equations Dx = (Vox)t and Dy = -(1/2)gt^2. The user successfully rearranged the first equation to express time (t = Dx/Vox) and substituted it into the second equation, leading to the expression Dy = -g(Dx^2/Vox^2)/2. A correction was provided regarding the proper placement of parentheses in the equation, confirming the user's approach as correct.

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Jeffeff
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Homework Statement


Using these two equations of (1) Dx=(Vox)t and (2) Dy= -(1/2)gt^2 I have to find an expression of Dy as a function of Dx

Vox = initial component of velocity at time =0
Dx = Distance traveled horizontally
Dy= Distance traveled vertically upward

Homework Equations


Dx=(Vox)t and Dy= -(1/2)gt^2


The Attempt at a Solution


I rearranged equation (1) to get time; t=Dx/Vox and used this in equation (2) to substitute for t and coming up with the equation Dy = -g(Dx^2/Vox^2/2. I'm not quite sure I did this right any help would be appreciated.

Thanks in advance!
 
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Jeffeff said:

Homework Statement


Using these two equations of (1) Dx=(Vox)t and (2) Dy= -(1/2)gt^2 I have to find an expression of Dy as a function of Dx

Vox = initial component of velocity at time =0
Dx = Distance traveled horizontally
Dy= Distance traveled vertically upward

Homework Equations


Dx=(Vox)t and Dy= -(1/2)gt^2

The Attempt at a Solution


I rearranged equation (1) to get time; t=Dx/Vox and used this in equation (2) to substitute for t and coming up with the equation Dy = -g(Dx^2/Vox^2/2. I'm not quite sure I did this right any help would be appreciated.

Thanks in advance!
Hello Jeffeff. Welcome to PF !

You have unbalanced parentheses. Assuming you mean that Dy = -g(Dx2/V0x2)/2, that's correct.
 
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