Skydiver Joseph Kittinger, record for the longest jump, reaches a terminal speed of 60m/s with his arms and body fully extended. Then, he switches to a head first dive, reducing his cross section area by 80% (that is, his cross section area when head-first is only 20% of his cross-section area when falling with arms and body extended). what is his terminal speed when falling head-first?
I know that I will be needing to use the drag force formula which is (1/4)Av^2. But have no idea how to start. Mass is not given...
thanks in advance
To find Joseph Kittinger's terminal speed when falling head-first, we can start by considering the drag force acting on him in both scenarios: with arms and body extended, and head-first.
The drag force formula is given by:
Drag force = (1/4) * A * v^2
Where:
Drag force is the force opposing the motion (in this case, the force of air resistance).
A is the cross-sectional area of the object.
v is the velocity of the object relative to the air.
Let's assume the cross-sectional area of Joseph Kittinger when his arms and body are extended as A1, and the cross-sectional area when he is head-first as A2.
Given that the cross-sectional area when head-first is only 20% of the cross-sectional area with arms and body extended, we can write:
A2 = 0.2 * A1
Now, let's consider the drag forces in each scenario separately.
With arms and body extended:
Drag force1 = (1/4) * A1 * v1^2
When he switches to a head-first dive, the drag force becomes:
Drag force2 = (1/4) * A2 * v2^2
We know that when his arms and body are extended, Joseph Kittinger reaches a terminal speed of 60 m/s. This means that the drag force and the gravitational force are equal, resulting in no net acceleration.
Therefore, we can write:
Drag force1 = Gravitational force
The gravitational force is given by:
Gravitational force = m * g
Where:
m is the mass of Joseph Kittinger.
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the mass is not given, we can consider it as a constant and can cancel it out when comparing the drag forces.
So we have:
(1/4) * A1 * v1^2 = m * g
Now, let's consider the head-first dive scenario.
The drag force2 when he is head-first is given by:
(1/4) * A2 * v2^2
Since the drag forces in both scenarios are equal to the gravitational force, we can write:
(1/4) * A1 * v1^2 = (1/4) * A2 * v2^2
Substituting the value of A2 from earlier, we have:
(1/4) * A1 * v1^2 = (1/4) * (0.2 * A1) * v2^2
Simplifying the equation, we get:
v2^2 = 5 * v1^2
Now, we know that v1 (the terminal speed with arms and body extended) is 60 m/s. Substituting this value, we have:
v2^2 = 5 * 60^2
Simplifying further, we get:
v2^2 = 5 * 3600
v2^2 = 18000
Finally, taking the square root of both sides, we find:
v2 = √18000
Thus, Joseph Kittinger's terminal speed when falling head-first is approximately 134.16 m/s.