- #1

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**above/below**where it was launched?

Thanks

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- Thread starter MattTuc13
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- #1

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Thanks

- #2

HallsofIvy

Science Advisor

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

x= v_{0}cos(θ)t, y= v_{0}sin(θ)t- (g/2)t^{2}.

Let y_{0} be the height of the landing point relative to the starting point (0 is same level, negative if below, positive if above). Solve

y= v_{0}sin(θ)t- (g/2)t^{2}= y_{0} 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 y_{0}= 0 but still a valid formula. Notice that when y_{0}= 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.

x= v

Let y

y= v

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 y

[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

Homework Helper

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The last equation (where y_{0} = 0) should be

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

(notice the v_{0}^{2})

Which simplifies to:

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

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

(notice the v

Which simplifies to:

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

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