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

Algebra ->  Real-numbers -> 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      Log On


   

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) About Me  (Show Source):
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+h+=+-16t%5E2+%2B+60t+%2B+14+
+h+=+30+
+30+=+-16t%5E2+%2B+60t+%2B+14+
+-16t%5E2+%2B+60t+-+16+=+0+
+-4t%5E2+%2B+15t+-+4+=+0+
+t+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+-4+
+b+=+15+
+c+=+-4+
+t+=+%28-15+%2B-+sqrt%28+15%5E2-4%2A%28-4%29%2A%28-4%29+%29%29%2F%282%2A%28-4%29%29+
+t+=+%28-15+%2B-+sqrt%28+225+-+64+%29%29%2F%28-8%29+
+t+=+%28-15+%2B-+sqrt%28+161+%29%29%2F%28-8%29+
+t+=+%28-15+%2B+12.689%29%2F%28-8%29+
+t+=+%28+-2.311+%29%2F%28-8%29+
+t+=+.289+
and
+t+=+%28+-15+-+12.689+%29%2F%28-8%29+
+t+=+%28+-27.689+%29%2F%28-8%29+
+t+=+3.461+
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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 )
----------------------
check:
Here's the plot:
+graph%28+400%2C+400%2C+-1%2C+6%2C+-8%2C+80%2C+-16x%5E2+%2B+60x+%2B+14+%29+
Looks like my answers are close -check the math