SOLUTION: The height of a cannonball shot upward from a height of 14 feet above the ground with an initial velocity of 60 feet/second is given by the equation
h equals negative 16 t squ
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Question 1143391: The height of a cannonball shot upward from a height of 14 feet above the ground with an initial velocity of 60 feet/second is given by the equation
h equals negative 16 t squared plus 60 t plus 14
h=−16t2+60t+14.
After how many seconds will the cannonball be 30 feet above the ground?
Found 2 solutions by Alan3354, josmiceli:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
duplicate
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
and
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It will be 30 ft above the ground at .289 sec and 3.461 sec
( first time going up, second time going down )
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check:
Here's the plot:
Looks like my answers are close -check the math
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