Projectile Motion

Example 1: A plane flying east at 300 m/s drops a package from an altitude of 5,000 meters. a)How long does it take to hit the ground? b)How far has it traveled horizontally? c)With what speed does it hit the ground?(this answer requires you to find V_ox and V_y(final) then use them to find the "resultant" velocity)

Projectile motion is characterized by the following four equations:

1. x = V_ox t

2. v_y = v_oy - g(t)

3. Dy = v_oyt - (1/2)g(t²)

4. v_y² = v_oy² - 2g(Dy)

We ignore air resistance to simplify the motions involved (and, unless you take fluid dynamics in college, you're likely not to calculate it). Note also that equations 2, 3, and 4 are for VERTICAL MOTION ONLY!!

So, to answer (a): Sometimes it pays to rephrase the question - how long does it take the object to fall from 5000 m?

Here's what we know:

Dy = 5,000 m

V_oy = 0 m/s

g = 9.80 m/s²

We can select equations 3 or 4 (both have Dy in them):

Let's try Dy = v_ot - (1/2)g(t²)

We know v_i is zero so we have:

Dy = 0 - (1/2)g(t²)

y - y_o = -1/2g(t²)

(0 - 5000m) / 4.9 m/s² = t²

t² = 1020.4 s²

t = 31.94 s

So that's how long it stays airborne. To answer part (b) we re-ask: how far can it travel horizontally in 31.94 s?

x = V_ox t

Since we know Vx already, this is just:

x = (300 /ms)(31.94 s)

x = 9583 m

Finally, we ask the question: what velocity does it hit the ground with? - this is a more involved question. Let's look at the graphics for this:

So we see that the projectile velocity into the ground can be resolved into a velocity in the y direction, V_y, and a velocity in the x direction, V_ox. These velocities are easily determined:

V_ox we already know because the horizontal velocity never changes (in our ideal, but unbreathable world). We can then find V_y by using one of the projectile equations for determining final velocity:

v_y = v_oy - g(t)

v_y = 0 - (9.8 m/s²)(31.94 s)

v_y = 313 m/s

Finally we revisit the old nemesis: vectors. Use the pythagorean theorem to determine V_f:

V_f = (v_y² + v_ox²)^1/2

V_f = (313² + 300²)^1/2

V_f = 433 m/s    (this is the "resultant")

Notice that it impacts the ground at a much higher velocity than if it just fell straight down…let that be a lesson to any parachutists out there.