Question

An 86.0 kg skydiver jumps out of a balloon at an altitude of 1 340 m and opens the parachute at an altitude of 160 m. (a) Assuming that the total retarding force on the diver is constant at 70.0 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? (b) Do you think the sky diver will be injured? Explain. (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 your answer.

Answer 1

a)We can consider this motion as two parts.

1.before the parachute is opened.

2.after opening the parachute.

For case 1 , down force = weight = mg = 86×9.8N

Retarding (or) upward force on the body=70N

Therefore net downward force, ma=(mg-70)

Which gives a=(mg-70)÷m=8.98m/s^2

Therefore V=√(u^2+2as)=√2as (since the diver jumped from a stationary balloon, the motion is free fall)

V=√(2×8.98×[1340-160]) = 145.57m/s

For case 2 , since retarding force is greater than the weight (or) upward force,

a=(3600-mg)÷m = 32.06m/s^2

V=√2as=√(2×32.06×160) = 101.28m/s(in the opposite direction)

Therefore velocity with which he reaches the ground=145.57-101.28=44.29m/s

b)The diver lands on the ground with a velocity of 44.29m/s which can cause severe injuries to him even death.

c)let 145.57-V=5m/s

Then V=140.57m/s

We have, V^2=2as

Therefore s=v^2÷2a

=(140.57^2)÷(2×32.06)

=308.171m

d)In reality, the retarding force is not constant but is proportional to its velocity.Since there is an acceleration due to gravity , velocity keeps increasing and proportional to it the force also increases to a limit (terminal velocity).So the retarding force is not constant.