Taking 'down' to be the positive direction, the acceleration of the object consists of two components, an acceleration of 9.8m/s^2 due to gravity and a retardation due to air resistance.

The retardation part is described as being 1.5m/s^2 per 7.5m/s of velocity, so if the velocity of the object is v, then the retardation part will be 1.5*(v/7.5) = v/5 m/s^2.

So, the acceleration of the object will be

\(\displaystyle a = \frac{dv}{dt}=9.8-\frac{v}{5}=\frac{49-v}{5}\).

Solving (separation of variables),

\(\displaystyle v=49-Ae^{-t/5}\), where \(\displaystyle A\) is an arbitrary constant.

For the constant to be evaluated, we should be told the initial velocity of the object. We are not, but I shall assume that the object was dropped from rest so that \(\displaystyle v=0 \text{ when } t=0\), otherwise ... .

Substituting , \(\displaystyle A = 49, \text{ so }v=49-49e^{-t/5}\).

\(\displaystyle v = \frac{ds}{dt}=49-49e^{-t/5}\),

so, integrating,

\(\displaystyle s = 49t+245e^{-t/5}+C\), where \(\displaystyle C\) is an arbitrary constant.

Taking the original drop point to be \(s=0, \text{ }t=0,\) substituting shows that \(C=-245\), so

\(s=49t+245e^{-t/5}-245\).

Then, if the time taken to hit the ground is 10.5 sec, the height h is given by

\(h= 514.5+245e^{-2.1}-245=299.5 \text{m}\).

 

The terminal velocity of the object (if it were allowed to continue falling) is \(\displaystyle \lim_{t \rightarrow \infty} v=49\) m/s.

 

Tiggsy

Hmmmm.... I do NOT know if any of these calculations are correct, but here is my try

v = vo + at      (1)

t=10.5 sec

+a = 9.81 m/s^2

-a = air resistance = 1.5 m/s^2  x  v m/s  /  7.5 m/s = .2 v      (2)

Substituting into equation (1):

v = 0 + (9.81 - .2v)(10.5)  = 103 - 2.1v

v+2.1v = 103

3.1v = 103     or v  = 33.225 m/s   at 10.5 seconds

 

Now let's find the height

x = xo + vot + 1/2 at^2           xo and vo are zero for this application, so we are left with

x = 1/2 a t^2     

x = 1/2 (9.81 - .2(33.225))(10.5)^2  = 174.47 m   (??)

                 

 

Terminal velocity is where air resistance EQUALS gravity

1.5m/s^2 x v m/s / 7.5m/s  = 9.81 m/s^2

.2 v = 9.81

v(terminal) = 9.81 /  .2  = 49.05  m/s

 

I am VERY unsure of ANY of this ...I hope someone will confirm it or laugh out loud at my answers!  (I'm almost afraid to post it!)

 

 

I have never answered a  question like this one either EP.

 

Don't be afraid to post things that you are unsure of.

 I am sure you have seen me make posts and state quite clearly that I am not sure if it is right.  

The asker always has the ability to accept or dismiss your answer. That is not our responsibility :))

It's a first order linear differential equation.

There are a variety of methods that could be used for its solution, integration factor or trial solution (look them up and try them)  both work, but I mentioned separation of variables in the earlier post so here's what that looks like.

 

As the name says, begin by separating the two variables v and t, by cross multiplication/division. 

We treat dv/dt as a fraction, which it isn't, but then again most of us turn a blind eye to that. (Beware the nitpicker !)

\(\displaystyle \frac{dv}{49-v}=\frac{dt}{5}\).

Quickly stick an integral sign on both sides (and then it's ok.)

\(\displaystyle \int \frac{1}{49-v}dv=\int \frac{1}{5}dt\),

and integrate

\(\displaystyle -\ln(49-v)=t/5 + C\).

Rearrange,

\(\displaystyle \ln(49-v)=-t/5-C\),

\(\displaystyle 49-v=e^{-t/5-C}=e^{-t/5}.e^{-C}=Ae^{-t/5}\),

( \(\displaystyle e^{-C} \) is just an arbitrary constant, it's convenient to replace it with a single letter).

Finally,

\(\displaystyle v=49-Ae^{-t/5}\).

 

BTW EP your attempt in #1 is incorrect because the equations you use are derived, by integration, from dv/dt = a, where the acceleration a is a constant. In this example the acceleration is not constant due to a variable air resistance.

 

Tiggsy.