Asked by MIKE
The distance in feet that a skydiver is from the ground (t) seconds after jumping from an airplane flying at 17,000 ft above ground is given by the following equation: d(t)=-16t^2 +17000. a.) how many feet above ground is the skydiver above ground after 4 seconds? b.)how many seconds have past when the skydiver reaches 3000 ft
Answers
Answered by
Reiny
a) just plug in 4 for t and evaluated,
remember to square first, that is,
-16(16) + 17000
b) set 3000 = -16t^2 + 17000
16t^2 = 14000
t^2 = ± √14000/4
= appr 29.6 sec. , obviously ignoring the negative.
(not realistic, skydivers do not fall according to that formula)
remember to square first, that is,
-16(16) + 17000
b) set 3000 = -16t^2 + 17000
16t^2 = 14000
t^2 = ± √14000/4
= appr 29.6 sec. , obviously ignoring the negative.
(not realistic, skydivers do not fall according to that formula)
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