# Factoring known drag into finding height of projectile

1. May 20, 2012

### Chark711

Alright, I am trying to work out some equations on a project to determine if fictional instances of physics are possible or not. In my case, I am seeing how a person travels if they are thrown from a jet that is traveling vertically (don't ask). I have completed the equations to solve how far the person would travel without air resistance, and they look like this:

Height initial=450 meters
Velocity initial=320 m/s
Time in air=6.5 seconds

Hf=Hi + Vi*T - .5gT^2
Hf=450 meters + 320 m/s * 6.5 s - (4.9m/s^2)6.5s^2
long story short, I get that the person is about 2,323 meters high after 6.5 seconds.

Now, how do I add drag? I did Force of drag equations and came to the concussion that the man experiences 3,763 newtons of drag. Is this reasonable? And does it factor into previous equations, or do I get a whole new equation I need to deal with now?

2. May 21, 2012

### LawrenceC

The situation is described by a differential equation. Have you studied math at that level? It also involves fluid mechanics and drag coefficients. The force impeding the upward motion is dependent on the square of velocity as well as gravity.