# Two Dimensional Motion

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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HallsofIvy
Homework Helper
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).
$$t= \frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}$$
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):
$$x= v_0cos(\theta)\frac{v_0sin(\theta)+\sqrt{v_0^2sin^2(\theta)-2gy_0}}{g}$$

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

I would consider it easier to solve the equations than to memorize that formula.

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Päällikkö
Homework Helper
The last equation (where y0 = 0) should be
$$x= \frac{2v_0^2sin(\theta)cos(\theta)}{g}$$
(notice the v02)
Which simplifies to:

$$x= \frac{v_0^2sin(2 \theta)}{g}$$

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