SOLUTION: A cannonball is shot into the air with a velocity of 60ft/sec. at what time will the cannonball first reach a height of 36ft, using the equation below? h=-16t^2 + 60t ???

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Question 772793: A cannonball is shot into the air with a velocity of 60ft/sec. at what time will the cannonball first reach a height of 36ft, using the equation below?
h=-16t^2 + 60t ???
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
h=-16t^2 + 60t
Plug in 36 for h, then simplify and solve that quadratic 
equation for t

You'll get two answers, t=3/4 and t=3.

It will be at a height of 36 feet first in 3/4ths of a second

It will also be at a height of 36 feet again in 3 seconds.

One answer is when the cannonball is going up, and the other 
answer is when it is coming down.

He wants the answer when it FIRST was at that height (when going up),
so it's 3/4 of a second.

Edwin
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