Nine lives are flashing before their eyes

First Convert miles per hour to meters per second

60 mph = 26.8224 mps

Now use the above formula...

Since the cat is falling with uniform velocity,

The gravitational force should be equal to the drag force

Thus 2 m g = A ( v ^ 2 )

=> 2 ( 5 ) ( 9.8 ) = A ( 26.8224 ^2 )

From this we get A = 0.136 square meters

I would suggest fixing the equation so that units make sense. In the end you do not come to an area in m^2, you get an answer of A=0.136kg/m... if you ignore units after getting all terms into meters and seconds it sort of works, but seriously the units should make sense in the end.

At terminal velocity, there is no vertical acceleration and that means that $F_d = W$ . Since $W = mg$ , we can safely say $F_d = 5\cdot 9.8 = 49N$ . We also have the value $v$ $= 60 mil/h$ since our gravitational acceleration is in $ms^{-2}$ , and the final answer has to be in $m^{2}$ , hence we have to convert $v$ from $mil/h$ to $ms^{-1}$ . Letting $1$ $mile = 1.6$ $km = 1600$ $m$ $60$ $mil/h$ $= \frac {60\times 1.6\times 1000}{60\times 60} ms^{-1} = 26.7$ $ms^{-1}$ . From the given formula we know that $F_d = \frac {1}{2}Av^{2}$ , which can be rewritten as $A = \frac {2F_d}{v^{2}}$ . Substituting in $F_d = 49$ $N$ and $v = 26.7$ $ms^{-1}$ , we get $A = \frac {2\times 49}{26.7^{2}} = 0.137$ $m^{2}$

When an object is falling at terminal velocity it experiences no acceleration so there are no resultant forces acting on the object. This means that the weight of the cat is equal to the force of drag on the cat at terminal velocity:

$5g = \frac{1}{2}Av^{2} [1]$

In order to work out the cross-sectional area in $m^{2}$ , the 60mph has to be converted into $ms^{-1}$ . This is done by first dividing the 60mph by 3600 to find it in miles per second and then multiplying by 1609.344 (conversion of miles to m) to find it in $ms^{-1}$ :

$60mph = (\frac{1}{60}*1609.344)ms^{-1} = 26.8224ms^{-1}$

Rearranging equation [1] for A, we get:

$A = \frac{10g}{v^{2}} = \frac{98}{719.44}$

By evaluating this you find that:

$A = 0.1362m^{2}$

A=(2F)/v^2 =2(5)(9.8m/s^2)/(26.817m/s)^2 = (98m/s^2)/(719.15m^2/s^2) = 0.136

F_d= m a = A (v)^2/2 => 5 9.8= A (711.11111111111111111111111111111)^2 since v=60 miles per second so 1miles = 1.6 km = 1600m so 60 miles per second = (1600*60)/3600 => A= 0.136 m^2 :D

At terminal velocity, the cat's drag force will equal the gravitational force. This allows the cat to maintain the same falling velocity, as no net force acts on it. Since the gravitational force is the acceleration due to gravity * the cat's mass, the gravitational force is 9.8 * 5 N = 49 N. This must equal the drag force, which is (A* v^2)/2. We are given that v = 60 mph, which me must convert to m/(s^2). (60 mi/ 1 hr) = (60 * 1600 m )/ (3600 s) which is about 26.822 m/s. Putting into our formula:

49 = (A * 26.822^2)/2, and by simple algebra, A=.139 m^2

It's simple to get the answer. Follow up the formula above: $Fd = \frac{1}{2}Av^{2}$ . $Fd$ in here, is the weight force of the cat, you can find it using the weight force as the mass of cat multiply the gravity of the place being. Finally, don't forget to change the conversion of velocity from miles per hours to meter per second . (Hint: 1 mile = 1.609,344 meters) .

At the terminal velocity, the overall acceleration is $0$ , i.e. the gravitational force is of the same magnitude as the drag force, but in the other direction, so

$F_g = F_d \implies mg = \frac{1}{2}Av^2 \implies A = \frac{2mg}{v^2} \approx \frac{2 \cdot 5 \cdot 9.8}{(60 \cdot 0.447)^2} \approx \boxed{0.136}\text{ m}^2.$

Note : $1 \text{ mph} \approx 0.447 \text{ m/s}.$