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Question 123536This question is from textbook Elementary and intermediate Algebra
: Skydiving. If there were no air resistance, then the height (in feet) above the earth for a skydiver t seconds after jumping from an airplane at 10,000 feet would be given by h(t)= -16t^2 + 10,000
a) find the time that it would take to fall to earth with no air resistance: that is, find t for which (t)= 0.A skydiveractually gets about twice as much free fall time due to air resistance.
b)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.
c) Is the skydiver's velocity increasing or decreasing as she falls?
This question is from textbook Elementary and intermediate Algebra
Found 2 solutions by stanbon, rajagopalan:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If there were no air resistance, then the height (in feet) above the earth for a skydiver t seconds after jumping from an airplane at 10,000 feet would be given by h(t)= -16t^2 + 10,000
a) find the time that it would take to fall to earth with no air resistance: that is, find t for which (t)= 0.A skydiveractually gets about twice as much free fall time due to air resistance.
When on earth the height is zero.
-16t^2+10,000=0
-16t^2 = -10,000
t^2 = 625
t = 25 seconds
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b)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.
Height after 5 seconds = h(5)=-16(5^2)+10000 = 9600 ft
The skydiver falls 400 ft in the 1st 5 seconds.
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Height after 20 seconds = h(20)=-16(20^2)+10000= 3600 ft
Height aftrer 25 seconds is zero
The skydiver falls 3600 ft in the last 5 seconds.
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c) Is the skydiver's velocity increasing or decreasing as she falls?
increasing.
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Cheers,
Stan H.
Answer by rajagopalan(174)  (Show Source):
You can put this solution on YOUR website! 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
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