that's why i can flip a quarter in the car without it hitting my face! - Unit4

Projectile motion took me such a long to understand, so here's my post for as far as i know!
To prove my understanding of projectile motion, i created my own word problem:

An airplane flying 1000 m/s dropped a parachuter and he reached his landing spot at sea level, 600 meters from where the plane was. How high up was the plane?


Answer, 17.64 meters above the ground.

How to solve this answer: First you have to write down the givens in a format showing the x and y axis. You must remember to follow the "vegas rule" (whatever happens on the x or y axis stays on the x or y axis respectively...). You should have a table of givens that look like this:


X-axis table explanation: the initial and final velocity is the same because the plane is not accelerating.  The plane isn't accelerating because, because it's velocity isn't changing.

Y-axis table explanation: the original velocity is zero m/s^2 because the parachuter was sitting in the plane, he wasn't moving and therefore didn't have any speed before jumping off. Acceleration is -9.8 m/s^2 because the parachuter is being pulled by gravity (-9.8 m/s^2) in a downward motion (according to the positive and negative is in arrow drawing.)

In order to find the height of the plane, we need to find the time it took first. Time is equal n the x-axis and y-axis, because we learned that objects travel at the same speed no matter the direction they are thrown at.

Finally, to solve the equation, use the DAT equation, or d=1/2at^2 + Vot using the givens from the x-axis.
You should get 600m = 1/2 (0m/s^2)t^2s + 1000t

600 = 1000t, t= 0.6
Once you get the time, you can plug it in the DAT equation again, but this time use the givens from the y-axis.
You should get Dm= 1/2 (-9.8m/s^2) 0.6^2s + 0t
 D=1/2(-4.9)(3.6)
D = 17.64 meters. 

This is not a picture of me parasailing. I'm parachuting.

The height of the plane was 17.64 meters when the parachuter jumped off!