SOLUTION: a stone is thrown straight upward with a speed of 20m/s,it is caught on its way down at a point 5.0m above where it was thrown,how far was it going when it was caught,and how long

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Question 490857: a stone is thrown straight upward with a speed of 20m/s,it is caught on its way down at a point 5.0m above where it was thrown,how far was it going when it was caught,and how long did the trip take
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
Initial velocity = 20
h(t) = -4.9t^2 +20t
Solve for t when h(t) = 5
--> -4.9t%5E2+%2B20t+=+5
--> -4.9t%5E2+%2B20t+-5+=+0
-----------------------------------
Use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29
---------------------------------
-->
--> t+=+%28-20+%2B-+sqrt%28302%29%29%2F%28-9.8%29
--> t+=+%2820+%2B-+sqrt%28302%29%29%2F9.8+
We want the time when the stone is on its way down which will be the longer time
--> t+=+%2820+%2B+sqrt%28302%29%29%2F9.8+=+3.814
Trip took 3.814 sec
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v(t) = -9.8t + 20
Solve for when t = 3.814
--> v+=+%28-9.8%29%283.814%29+%2B+20+=+-17.378
Speed is always positive
Stone was traveling at 17.378 m/s when caught