Calculate Horiz. Dist. of 2D Object at Angle w/ V0 & y

[MattTuc13]

Is there an equation that gives the horizontal distance traveled 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

 

[HallsofIvy]

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.

 

[Päällikkö]

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]