SOLUTION: the distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the ground). If an object fell 63 feet in 8 seconds, how
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Question 761451: the distance that a free falling object falls is directly proportional to the square of the time it falls (before it hits the ground). If an object fell 63 feet in 8 seconds, how farr will it have fallen by the end of 10 seconds? ( leave the variation constant in fraction form or round to at least 2 decimal places. Round your final answer to the nearest foot.)
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Let d = distance fallen in feet and
Let t = fall time in seconds and
Let c = constant of direct proportionality in feet per seconds squared.
The formula to use is
(1) d = c*t^2
Where we are given
(2) 63 = c*8^2 or
(3) c = 63/64
Then in 10 seconds the distance is given by
(4) d = 63/64*(10)^2 or
(5) d = 6300/64 or
(6) d = 98.4375
Answer: the constant of variation is 63/64 and the distance dropped in ten seconds is 98 feet.
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