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
!
