# Integrating to find distance and time - difficult!

1. Apr 23, 2010

### 1234john

I have been looking for problems to work out for practice and came across one I have no idea how to even start about.

Say a box is dropped from a plane, with a parachute attached to it. If we are given two equations for velocity in terms of time, such as

Vx(t)=Vo cos(θ) e^(-pt) and Vy(t)=Vo sin(θ) - ge^(-z(t+6)) -16

If p is the drag coefficient in the x and z is the drag coefficient in the y.
θ would be the angle between the ground and the box.

Would θ not be a constant though since the angle will always be constant, even if the box follows a parabolic path?

And say the plane was flying level when the box was dropped... Wouldn't this mean that there is only an initial velocity in the x direction at whatever speed the plane was flying?

For practice, I made up p=0.061 and z=0.213

My whole dilemma is how I would find the time it takes for the box to hit the ground.

I'm fairly sure I'd have to take the integral which would give me distance. Plugging in zero for Vo in the y direction would cancel out the first part of the equation. θ would remain constant which would take out the sin and cos part of each equation and make it a constant.

I guess from there I am lost. My above assumptions could be wrong, however.

Please let me know! No rush on answering this as its just for my own personal practice and "fun".

2. Apr 23, 2010

### reckon

as Vo is y direction is zero
Vy(t)=-g exp(-z(t+6))-16

Vy=dy/dt, it makes
dy= (-g exp(-z(t+6))-16 )dt

and you get what you want