An 80.0 kg skydiver jumps out of a balloon at an altitude of 1.00 x 10 m and opens the parachute at an altitude
of 200.0 m. omitior store it in any other
a) Assuming that the total retarding force on the diver is constant at 50 N with the parachute closed and constant
at 3.60 x 10^Nwith the parachute open, what is the speed of the diver when he lands on the ground?
b) think ? .
c) At what height should the parachute be opened so that the final speed of the skydiver when he hits the ground
is 5.00 m/s?
d) How realistic is the assumption that the total retarding force is constant?
Explain.
a) To find the speed of the diver when he lands on the ground, we can use the concept of work-energy theorem. The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. We can calculate the work done on the skydiver using the formula:
Work = Force * Distance
Since the retarding force is constant, the work done by the force can be found by multiplying the force by the distance traveled. Initially, the skydiver jumps out of the balloon at an altitude of 1.00 x 10 m. So the work done without the parachute is:
Work without parachute = 50 N * (1.00 x 10 m)
When the parachute is opened, the retarding force becomes 3.60 x 10^N. The skydiver then descends from 1.00 x 10 m to the ground level (0 m). So the work done with the parachute is:
Work with parachute = 3.60 x 10 N * (1.00 x 10 m - 0 m)
We can now equate the work done to the change in kinetic energy of the skydiver:
Work without parachute + Work with parachute = 1/2 * mass * (final velocity)^2
Since the mass of the skydiver is 80.0 kg, we can substitute the values and solve for the final velocity.
b) It seems like there is a missing question or unclear statement. Could you please rephrase or provide more information to assist you better?
c) To find the height at which the parachute should be opened so that the final speed of the skydiver when he hits the ground is 5.00 m/s, we can use a similar approach as in part a.
In this case, the work done without the parachute will remain the same as before since the retarding force is constant at 50 N. However, we have a different work done with the parachute. Let's denote the height at which the parachute is opened as 'h'. The work done with the parachute can be calculated as:
Work with parachute = 3.60 x 10 N * (h - 0)
Again, we equate the work done to the change in kinetic energy:
Work without parachute + Work with parachute = 1/2 * mass * (final velocity)^2
Substituting the known values, including the mass of the skydiver as 80.0 kg and the final velocity as 5.00 m/s, we can solve for the height 'h'.
d) The assumption that the total retarding force is constant may not be entirely realistic in this scenario. In reality, the total retarding force experienced by the skydiver would depend on various factors such as air resistance, wind speed, parachute efficiency, and the diver's body position.
For example, air resistance increases with the speed of the skydiver, which would result in a higher retarding force as the skydiver gains velocity. Additionally, the angle at which the body is aligned with the airflow can affect the retarding force. Therefore, the assumption of a constant retarding force is simplified for the purpose of this problem and does not accurately represent the real-world scenario.
1000 meters maybe ? 3600 N maybe???
If so
before chute opens:
Vi = initial speed = 0
g = -9.81 m/s^2
Force up on man = 50 N
acceleration down = 9.81-50/80 = 9.2 m/s^2
v = 0 + 9.2 t
h = 1000 - (9.2/2) t^2
200 = 1000 - 4.6 t^2
t^2 = 174
t = 13.2 seconds at open chute at 200 meters, v = 9.2 t = 121 m/s
Now 200 meters more with chute open and initial speed =121 m/s
F = m g - 3600
a = g - 360/80 = 9.81 - 45 = - 35.2
so
v = 121 - 35.2 t downward
h = 0 = 200 - (1/2)(35.5) t^2
17.6 t^2 = 200
t = 3.37 seconds
v = 121 - 35.2*3.37 = 121 - 119 = 3 m/s
You can do the rest. Check my arithmetic!
for part b, drag is sort of proportional to v^2, not constant