Hi All,

I need help with this question badly. I have am stuck on this one part, so I cannot finish the question.

Here it is:

"A 1-tonne(2000)kg object is launched vertically with an initial velocity of 30,000m/s at time t=0 and at height y=0. The only forces working on the object are gravity and air friction. (i.e. the object will land where it was launched from). Considering such high speeds and large distances are involved, the model must incorporate air friction and variations in gravity.Air resistance is a force that is proportional to the square of the speed, while gravity follows an inverse square law.

A difficulty is that the air resistance co-efficient varies with the hight of the object, because the atmospheres density changes

The density at y=0 is 1.275kg/m^3.

The density at y=6000m is half the density at y=0.

There was some more information given in the question, which enabled me to determine the air density for any height,y. Which is(I think)

D(y) = 1.275e^(-(ln2/6000)*y)

Also, I have data regarding the drag force for certain speeds at certain air densities, which I have given to you in a jpeg below.

My task is to determine how high the object will go, and what is the time taken to return to earth.

I know that the co-efficient of air-resistance is a function of air density, and that the frictional co-efficient is related to air density by a quadratic relationship. My first question is: how do I determine this relationship from the data given, and once I have that (lets call it k), would the formula for Drag Force, be F(v) = kv^2, where v=velocity of the object?

My second question is about the variations in gravity. Can ayone explain to me how the inverse-square law would apply to this scenario? I do have a model for the variations in gravity, but I dont think it is right. What formula would I use in determining the force opposing the direction of the initial velocity (i.e. -mg, since gravity isn't constant, what formula do I use for g?)

If anybody could help me, I would really appreciate it.

Thank you