An 82.0 kg skydiver jumps out of a balloon at an altitude of 1 500 m and opens the parachute at an altitude of 195 m.
(a) Assuming that the total retarding force on the diver is constant at 55.0 N with the parachute closed and constant at 3 600 N with the parachute open, what is the speed of the diver when he lands on the ground?
To find the speed of the skydiver when he lands on the ground, we need to calculate the potential energy lost and the work done by the retarding force.
Step 1: Calculate the potential energy lost:
The potential energy lost by the skydiver is equal to the change in potential energy between the starting altitude and the landing altitude.
ΔPE = m * g * Δh
Where:
m = mass of the skydiver = 82.0 kg
g = acceleration due to gravity = 9.8 m/s^2
Δh = change in altitude = (1,500 m - 195 m) = 1,305 m
Substituting the values:
ΔPE = 82.0 kg * 9.8 m/s^2 * 1,305 m
ΔPE = 1,032,372 J
Step 2: Calculate the work done by the retarding force:
When the parachute is closed, the retarding force is 55.0 N, and when the parachute is open, it is 3,600 N. We need to calculate the work done in both cases.
Work done with parachute closed:
W1 = F1 * d1
Where:
F1 = retarding force with parachute closed = 55.0 N
d1 = distance traveled with parachute closed = (1,500 m - 195 m) = 1,305 m
Substituting the values:
W1 = 55.0 N * 1,305 m
W1 = 71,775 J
Work done with parachute open:
W2 = F2 * d2
Where:
F2 = retarding force with parachute open = 3,600 N
d2 = distance traveled with parachute open = (195 m - 0 m) = 195 m
Substituting the values:
W2 = 3,600 N * 195 m
W2 = 702,000 J
Step 3: Calculate the kinetic energy at landing:
The change in kinetic energy is equal to the work done by the retarding force minus the potential energy lost.
ΔKE = W2 + W1 - ΔPE
Where:
W2 = work done with parachute open = 702,000 J
W1 = work done with parachute closed = 71,775 J
ΔPE = potential energy lost = 1,032,372 J
Substituting the values:
ΔKE = 702,000 J + 71,775 J - 1,032,372 J
ΔKE = -258,597 J (negative value indicates a decrease in kinetic energy)
Step 4: Calculate the speed at landing:
The kinetic energy at landing is equal to (1/2) * m * v^2, where v is the landing speed.
(1/2) * m * v^2 = ΔKE
v^2 = (2 * ΔKE) / m
v = √(2 * ΔKE / m)
Substituting the values:
v = √(2 * (-258,597 J) / 82.0 kg)
v = √(-6295.4 m^2/s^2)
v = -79.37 m/s
Since speed cannot be negative, the magnitude of the speed at landing is 79.37 m/s.
To find the speed of the skydiver when he lands on the ground, we can use the concept of mechanical energy conservation.
Let's start by calculating the gravitational potential energy of the skydiver at an altitude of 1,500 m:
Potential energy (PE) = mass (m) x acceleration due to gravity (g) x height (h)
The mass of the skydiver is given as 82.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2. Therefore:
PE1 = 82.0 kg x 9.8 m/s^2 x 1500 m
Next, let's calculate the potential energy of the skydiver at an altitude of 195 m:
PE2 = 82.0 kg x 9.8 m/s^2 x 195 m
Since the retarding force on the diver is constant, the work done by this force is given by the formula:
Work = Force x Distance
So, we need to calculate the work done with the parachute closed and open.
First, for when the parachute is closed:
Work1 = retarding force with parachute closed x (1500 m - 195 m)
Next, for when the parachute is open:
Work2 = retarding force with parachute open x 195 m
According to the principle of mechanical energy conservation, the change in potential energy is equal to the work done by external forces. Therefore:
PE1 - PE2 = Work1 + Work2
Now, we can substitute the known values and solve for the change in potential energy:
82.0 kg x 9.8 m/s^2 x 1500 m - 82.0 kg x 9.8 m/s^2 x 195 m = 55.0 N x (1500 m - 195 m) + 3600 N x 195 m
Simplifying the equation gives us:
Change in potential energy = 35745 J
Now, let's calculate the final kinetic energy of the skydiver when he lands on the ground:
Final kinetic energy (KEf) = Change in potential energy
KEf = 35745 J
The formula for kinetic energy is:
KE = 1/2 x mass x velocity^2
We can rearrange the formula to solve for velocity:
velocity = sqrt((2 x kinetic energy) / mass)
Substituting the known values:
velocity = sqrt((2 x 35745 J) / 82.0 kg)
Calculating the square root:
velocity = sqrt(871.71 m^2/s^2)
So, the speed of the diver when he lands on the ground is approximately:
velocity ≈ 29.56 m/s