An object launched at a known speed, but unknown angle

Question

An object is launched from a height of 1,000 m, at a speed of 500 m/s. If we want the object to land 14,000 m downrange, what angle (to two decimals) does the object need to fired at?

Answer 1

Let's use "θ" to represent our unknown angle. Now let's write down everything we do know.

initial velocity => vi = 500 m/s

horizontal displacement => Δx = 14,000 m

horizontal acceleration => ax = 0

vertical displacement => Δy = -1,000 m

vertical acceleration => ay = g = -9.8 m/s^2

In physics, we need to approach the horizontal (x) and vertical (y) trajectories separately, each with its own velocity, acceleration, and displacement. The only thing that is shared between our x and y planes is time (t). This means that if we solve for time (t) using our y-values, we will also be able to use that time (t) for calculations in the x direction.

Our initial velocity (vi) is composed of an x and y component (vi_x and vi_y). We must separate these components so that we can use them in our respective calculations. This is where our angle θ comes in.

We find that

vi_x = vi cos(θ) = 500cos(θ)

vi_y = vi sin(θ) = 500sin(θ)

We can now use one of our kinematics equations. Let's start with the horizontal trajectory. We want to choose an equation that uses both of our knowns (Δx and ax) as well as our unknown (vi_x). We do not want the final velocity (vf) to be a pert of the equation since it is not given, and we do not need to solve for it. This leads us to choose the following equation:

Δx = vi_x t + 1/2 ax t²

Unfortunately, this equation does also include t, a variable we do not know. However, since t is the same for both x and y, we can solve for it by setting up a separate equation in the y direction.

Δy = vi_y t + 1/2 ay t²

Now we can substitute our values for the variables we know in each equation. For x, that gives us:

14,000 = 500cos(θ) t + 1/2 (0) t^2

14,000 = 500cos(θ) t

For y, we get:

-1,000 = 500sin(θ) t + 1/2 (-9.8) t^2

Now we have 2 equations and 2 unknowns, so we can use our calculator to solve for both t and θ!