Parametric equation that describes Motion

Question

Write a set of parametric equations to describe the motion of a golf ball that is hit with a velocity of 105 ft/s at an angle of 30°.

Answer 1

For a projectile like this, the whole idea is that we split the motion into the horizontal piece and the vertical piece and let each one have its own equation, since they behave completely differently. Horizontally nothing slows the ball down in this model, so it just travels at a constant speed. Vertically gravity is pulling on it the entire time, so that one picks up a squared term.

First we need to break the 105 ft/s into its components using right triangle trigonometry, where the velocity is the hypotenuse and the 30° is the launch angle:

v_x = 105cos30° = 105(√3/2) ≈ 90.93 ft/s

v_y = 105sin30° = 105(1/2) = 52.5 ft/s

If you notice, the cosine gives the horizontal piece and the sine gives the vertical piece, and that is just SOH CAH TOA applied to the velocity triangle. Now the horizontal equation is distance equals rate times time, and the vertical equation is the same thing with the gravity term subtracted off, which in feet is 16t². So our parametric equations are:

x(t) = 105cos30° t, or x(t) = (105√3/2)t

y(t) = 105sin30° t - 16t², or y(t) = 52.5t - 16t²

That pair right there is the answer to what was asked. We are assuming the ball starts at ground level, so there is no constant added onto y. If your problem says it was teed up some number of feet off the ground, you would just add that number to the end of y(t).

Remember, the reason we want it in parametric form instead of as one equation is that t lets us answer questions about when, not just about shape. For example, the ball is at its highest when the vertical part stops rising, and since y is a downward parabola we can find that with the vertex formula from Algebra 2:

t = 52.5/32 = 1.640625 seconds, and y at that time is 43.07 feet

And it lands when y goes back to zero, so we set 52.5t - 16t² = 0 and factor out a t to get t(52.5 - 16t) = 0, which gives t = 0 at the start and:

t = 52.5/16 = 3.28125 seconds

Notice that the landing time is exactly twice the time to the top, which makes sense because the path is symmetric. Plugging that into x gives a distance of about 298.4 feet, so a little under 100 yards.

One last thing that is good to know, if you ever need the regular y in terms of x version, you solve the x equation for t and substitute it into y. You get a downward parabola, which is the actual shape of the path, and the parametric form is just a nicer way to carry the timing information along with it.

So, our final answer is x(t) = 105cos30° t and y(t) = 105sin30° t - 16t².