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 can break down the problem into two parts: the free fall with the parachute closed and the descent with the parachute open.
Step 1: Free fall with the parachute closed
During the free fall, the only force acting on the skydiver is gravity. We can use the equations of motion to determine the speed of the skydiver when the parachute opens.
Using the equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can solve for the initial velocity (u).
Given:
Mass of the skydiver (m) = 82.0 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Distance traveled during free fall (s) = 1500 m
Force during free fall (F) = 550 N (retarding force with the parachute closed)
The force acting on the skydiver during free fall is equal to the product of the mass and acceleration due to gravity (F = mg). Therefore, we can write down the equation: F = ma.
Rearranging the equation to solve for acceleration:
a = F/m
Substituting the values:
a = 550 N / 82.0 kg = 6.707 m/s^2
Now, we can calculate the initial velocity (u) by substituting the known values into the equation: v^2 = u^2 + 2as.
Rearranging the equation to solve for the initial velocity:
u = √(v^2 - 2as)
Substituting the values:
u = √(0 - 2 * 6.707 m/s^2 * 1500 m) = √(0 - 20121) ≈ -141.9 m/s
Note: The negative sign indicates that the initial velocity is in the opposite direction of the final velocity.
Step 2: Descent with the parachute open
After the parachute opens, there is an additional force acting on the skydiver, causing a change in acceleration. We can calculate the final velocity after descent using the same equation of motion approach.
Given:
Force during descent (F) = 3600 N (retarding force with the parachute open)
Distance traveled during descent (s) = 195 m
When the parachute is open, the total force acting on the skydiver is the sum of the retarding force and the force due to gravity.
Total force (F_total) = Force due to gravity (mg) + Retarding force (F_retarding)
Rearranging the equation to solve for acceleration:
a = (F_total) / m
Finding the force due to gravity:
Force due to gravity = mg
Substituting the values:
Force due to gravity = 82 kg * 9.8 m/s^2 = 803.6 N
Total force = Force due to gravity + Retarding force = 803.6 N + 3600 N = 4403.6 N
Now, we can calculate the final velocity (v) by substituting the known values into the equation: v^2 = u^2 + 2as.
Rearranging the equation to solve for the final velocity:
v = √(u^2 + 2as)
Substituting the values:
v = √((-141.9 m/s)^2 + 2 * 4403.6 N / 82 kg * 195 m) ≈ 39.7 m/s
Therefore, the speed of the skydiver when he lands on the ground is approximately 39.7 m/s.