Question

A skydiver weighing 588 N reaches a velocity of 45 m/s [down] before opening her parachute. After falling an additional 30 m in 4 seconds, her velocity has decreased to 25 m/s.

What is the acceleration of the skydiver over the 30 m displacement?

What was the upward force exerted on the parachute during that time?

Answer 1

The acceleration of the skydiver assuming that it is constant due to the parachute is -5
`m/s^(2)`.

The upwards force exerted during that time is 888 N.

Step-by-step explanation:

The skydiver weighs 588 N , this means that mg=588 N, where m is the mass of the skydiver.

Initial Velocity(u)=45 m/s

Final Velocity(v)=25 m/s

Time elapsed=4 s

a=??

Applying first equation of motion,

`v=u+at`

25=45 + 4a

4a=-20

a=-5
`m/s^(2)`

The acceleration of the skydiver assuming that it is constant due to the parachute is -5
`m/s^(2)`.

From the FBD attached we get that

F-mg=m x a

F-588=300

F=888 N

The upwards force exerted during that time is 888 N.