Question 123536
Part aaa.
Skydiving. 
height (in feet) above the earth=h(t)= -16t^2 + 10,000
the time it would take to fall when  h is 0 
-16t^2 + 10,000=0
-16t^2=-10000
16t^2=10000
t^2=10000/16=625
t=sqrt 625=25 seconds
okay

Part bbb)Use the accompanying graph to determine whether the skydiver(with no air resistance ) falls farther in the first 5 seconds or the last 5 seconds of the fall.

Though no graph is given we can calculate this:
Fall distance in first 5 sec =-16t^2 =-16(5*5)=-16*25=-400 m =400 m 
Fall distance in  Last 5 sec =ht after 25 sec-ht after 20 sec
ht after 25 sec=-16t^2 + 10,000=-16(25*25)+10000=-10000+10000=0
ht after 20 sec=-16t^2 + 10,000=-16(20*20)+10000=- 6400+10000=3600 m

So in the Last 5 secs he falls thro 3600 m as against 400m in the first 5 secs.


Part ccc) Is the skydiver's velocity increasing or decreasing as she falls?

Mean Velocity in first 5 secs =dist/time=400/5=80 m/sec
Mean Velocity in Last  5 secs =dist/time=3600/5=720 m/sec
Obviously Velocity is increasing.

In fact he is accelerating.
Note : Ignore the signs for distance as descending is shown as negative.
Cheers
good day