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(13215)   (Show Source): You can put this solution on YOUR website!
 
NOTE: The "^" symbol (shift-6) is often used to denote exponents....
To find the time when the cannonball is 30 feet above the ground, set h(t)=30 and solve:
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. 
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