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?
121.24 m/s
To find the speed of the diver when he lands on the ground, we need to calculate the energy changes at different altitudes.
1. Find the initial total mechanical energy at the higher altitude (1,500 m):
The initial total mechanical energy of the diver can be calculated using the equation:
Einitial = mgh
where m is the mass of the skydiver (82.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the initial altitude (1,500 m).
Einitial = (82.0 kg)(9.8 m/s^2)(1,500 m)
Einitial = 1,173,600 J
2. Find the total mechanical energy at the lower altitude (195 m) with the parachute closed:
The total mechanical energy at the lower altitude with the parachute closed can be calculated using the equation:
Eclosed = mgh + Wclosed
where Wclosed is the work done by the retarding force (55.0 N) against gravity over the distance discussed (1,305 m).
Wclosed = Fclosed * d
where Fclosed is the retarding force with the parachute closed (55.0 N) and d is the distance the parachute is closed (1,305 m).
Wclosed = (55.0 N)(1,305 m)
Wclosed = 71,775 J
Eclosed = (82.0 kg)(9.8 m/s^2)(195 m) + 71,775 J
Eclosed = 158,310 J + 71,775 J
Eclosed = 230,085 J
3. Find the total mechanical energy at the lower altitude (195 m) with the parachute open:
The total mechanical energy at the lower altitude with the parachute open can be calculated using the equation:
Eopen = mgh + Wopen
where Wopen is the work done by the retarding force (3,600 N) against gravity over the distance discussed (1,305 m).
Wopen = Fopen * d
where Fopen is the retarding force with the parachute open (3,600 N) and d is the distance the parachute is open (1,305 m).
Wopen = (3,600 N)(1,305 m)
Wopen = 4,713,600 J
Eopen = (82.0 kg)(9.8 m/s^2)(195 m) + 4,713,600 J
Eopen = 158,310 J + 4,713,600 J
Eopen = 4,871,910 J
4. Find the velocity of the skydiver when he lands on the ground (altitude = 0 m):
The final total mechanical energy at the ground level can be calculated using the equation:
Efinal = mghfinal
where hfinal is the final altitude (0 m).
Efinal = (82.0 kg)(9.8 m/s^2)(0 m)
Efinal = 0 J
Since energy is conserved, the initial energy equals the final energy at the ground level:
Einitial = Efinal
1,173,600 J = Eopen
Now we can calculate the final velocity of the skydiver using the equation:
Eopen = 0.5mv^2
where v is the final velocity.
4,871,910 J = 0.5(82.0 kg)v^2
9,743,820 J = 82.0 kg v^2
v^2 = 9,743,820 J / (82.0 kg)
v^2 = 118,876.098 m^2/s^2
v = √(118,876.098 m^2/s^2)
v = 344.586 m/s
Therefore, the speed of the diver when he lands on the ground is approximately 344.6 m/s.