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 1143395: 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?
Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!


NOTE: The "^" symbol (shift-6) is often used to denote exponents....

h%28t%29+=+-16t%5E2%2B60t%2B14

To find the time when the cannonball is 30 feet above the ground, set h(t)=30 and solve:

-16t%5E2%2B60t%2B14+=+30
-16t%5E2%2B60t-16+=+0
-4t%5E2%2B15t-4+=+0

It doesn't look like that's going to factor, so solve using the quadratic formula or a graphing calculator.

Note that the cannonball starts below 30 feet, so there will be two times when it is at 30 feet -- one when it is rising and another when it is falling.