Askydiver 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?

The acceleration of the skydiver assuming that it is constant due to the parachute is -5.

The upwards force exerted during that time is 888 N.

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,

25=45 + 4a

4a=-20

a=-5

The acceleration of the skydiver assuming that it is constant due to the parachute is -5.

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.

f=gm1m2/d^2 it d

f is force g is gravitaiton constant, m1 is mass of first object, m2 is mass of second object. d is distance between two object

d they tend to lose electrons and become cations.

explanation: