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Answer:
- 168.25 ft
- 3.125 s
Step-by-step explanation:
It is almost too easy to let a graphing calculator show you the solution. (See attached)
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You may be expected to use some actual algebra (or even calculus) to find the solution.
If you've spent any time with quadratics, you know the vertex of ax²+bx+c is located at x=-b/(2a). Here, that means the time to get to the highest point is ...
... t = -100/(2·(-16)) = 100/32 = 3.125 . . . . seconds
Then the highest point is ...
... h(3.125) = -16·3.125² +100·3.125 +12 = 168.25 . . . . feet
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Or, you can put the equation into vertex form.
... h(t) = -16t² +100t +12
... = -16(t² +6.25t) +12
... = -16(t² +6.25t +3.125²) +12 +16·3.125² . . . . . add the square of half the t coefficient inside parentheses; add the opposite of the same amount outside parentheses
... = -16(t +3.125)² +168.25 . . . . . simplify
This tells us the peak of the travel of the bullet is at 168.25 ft after 3.125 seconds.
if the number in the parentheses is less than 1, is not growth but Decay, thus a negative rate, or namely is decreasing by that much per cycle.