Two Dimensional Motion

  • Thread starter MattTuc13
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  • #1
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Is there an equation that gives the horizontal distance travelled of an object projected at an angle (theta) with an initial velocity (V0) and given that the object lands (y) above/below where it was launched?

Thanks
 

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  • #2
HallsofIvy
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Of course. Just use the same equations you would if the landing point were on the same level:

x= v0cos(θ)t, y= v0sin(θ)t- (g/2)t2.

Let y0 be the height of the landing point relative to the starting point (0 is same level, negative if below, positive if above). Solve
y= v0sin(θ)t- (g/2)t2= y0 for t. (You will get two solutions the time for the landing is the larger of the two).
Since that is a quadratic equation, use the quadratic formula:
[tex]t= \frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]
Now put that value of t in the equation for x (use the positive sign for the root to get the larger of the two solutions):
[tex]x= v_0cos(\theta)\frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}[/tex]

A bit more complicated than the case y0= 0 but still a valid formula. Notice that when y0= 0 that reduces to
[tex]x= \frac{2v_0sin(\theta)cos(\theta)}{g}[/tex]
the usual formula.

I would consider it easier to solve the equations than to memorize that formula.
 
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  • #3
Päällikkö
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The last equation (where y0 = 0) should be
[tex]x= \frac{2v_0^2sin(\theta)cos(\theta)}{g}[/tex]
(notice the v02)
Which simplifies to:

[tex]x= \frac{v_0^2sin(2 \theta)}{g}[/tex]
 
Last edited:

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