# Thread: A mass projected vertically upward.

1. ## A mass projected vertically upward.

A body of mass 7 kg is projected vertically upward with an initial velocity 32 meters per second.

The gravitational constant is g = 9.8 m/s^2. The air resistance is equal to k*abs(v) where k is a constant.

Find a formula for the velocity at any time ( in terms of k):

v(t) = ?

I'm not really sure how to go about starting this problem. So any help i can get would be appreciated thanks!

2. ## Re: A mass projected vertically upward.

Originally Posted by bwat687
A body of mass 7 kg is projected vertically upward with an initial velocity 32 meters per second.

The gravitational constant is g = 9.8 m/s^2. The air resistance is equal to k*abs(v) where k is a constant.

Find a formula for the velocity at any time ( in terms of k):

v(t) = ?
going up

dv/dt = - 9.8 - kv

going down

dv/dt = 9.8 - kv

Now continue...

I'm not really sure how to go about starting this problem. So any help i can get would be appreciated thanks!

3. ## Re: A mass projected vertically upward.

Alright so i start with dv/dt = - 9.8(m) - kv m = mass

From there is this right.

dv/dt + (1/k)v = -9.8(m)

Then i found an integrating factor mew(t) = exp of the integral 1/k dt = e^(t/k)

integral of d/dt(e^(t/k)v) = integral of -9.8(m)e^(t/k) dt

e^(t/k)v = -9.8k(m)e^(t/k) + c

then solve for c using the initial condition of v(0) equals 32m/s

If this isn't right can you give me some direction as to where i went wrong thanks!

4. ## Re: A mass projected vertically upward.

The air resistance is equal to k*abs(v) where k is a constant.
bwat687,

Just as an aside, you might want to check your problem statement. Drag is more closely approximated by the square of the velocity. Just a thought...

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