SOLUTION: At 3:10P.M. a skydiver jumps from an airplane at an altitude of 15,000 feet. At 3:10:26 at an altitude of 4200, the skydiver opens the parachute. What was the skydiver's average sp

Question 100604This question is from textbook
: At 3:10P.M. a skydiver jumps from an airplane at an altitude of 15,000 feet. At 3:10:26 at an altitude of 4200, the skydiver opens the parachute. What was the skydiver's average speed during the 26-second free-fall? This question is from textbook

Answer by doukungfoo(195) (Show Source): You can put this solution on YOUR website!
This problem deals with distance, speed and time.
to find average speed you must divide distance by time.
So first we need to find the distance traveled by the skydiver
during his/her freefall.
The skydiver started at an altitude of 15,000 feet and opened the parachute
at 4200 feet.
So the distance traveled during freefall is
15000 - 4200 = 10800 feet
Now just divide the distance by the time
10800 ft / 26 seconds = average speed
415.3846154 ft/s = average speed
Note: since the distance unit given is feet and the time unit given
is seconds. The answer is expressed in feet per second. (ft/s)
Also you can round the answer or not depending if specified by your
instructor or text book.
For example the answer rounded to the nearest whole foot would be:
415 ft/s

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