Projectile motion equation help

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The discussion revolves around deriving an expression for vertical distance (Dy) as a function of horizontal distance (Dx) using the equations of projectile motion. The user correctly rearranged the first equation to solve for time (t) and substituted it into the second equation. The resulting expression is Dy = -g(Dx^2/Vox^2)/2, which is confirmed to be correct, although there was a note about unbalanced parentheses. The conversation emphasizes the importance of proper notation in mathematical expressions. Overall, the user receives validation for their solution approach.
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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