Answer: 2 < t < 3
This is between 2 and 3 seconds, excluding both endpoints.
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Explanation:
t = elapsed time in seconds
h(t) = height of the ball at time t
Replace h(t) with 100. Then get everything to one side.
[tex]h(t)= -16t^2+80t+4\\\\100= -16t^2+80t+4\\\\0 = -16t^2+80t+4-100\\\\0= -16t^2+80t-96\\\\-16t^2+80t-96 = 0[/tex]
Next, we'll use the quadratic formula to isolate t.
Plug in a = -16, b = 80, c = -96 which are the coefficients of the last equation mentioned above.
[tex]t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\t = \frac{-80\pm\sqrt{(80)^2-4(-16)(-96)}}{2(-16)}\\\\t = \frac{-80\pm\sqrt{256}}{-32}\\\\t = \frac{-80\pm16}{-32}\\\\t = \frac{-80+16}{-32} \ \text{ or } \ t = \frac{-80-16}{-32}\\\\t = \frac{-64}{-32} \ \text{ or } \ t = \frac{-96}{-32}\\\\t = 2 \ \text{ or } \ t = 3\\\\[/tex]
Therefore, the ball's height is greater than 100 ft in the air for the interval [tex]2 < t < 3[/tex]
This is between 2 and 3 seconds, excluding both endpoints.
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Check:
Let's plug in t = 2 to find that...
[tex]h(t)= -16t^2+80t+4\\\\h(2)= -16(2)^2+80(2)+4\\\\h(2)= -16(4)+80(2)+4\\\\h(2)= -64+160+4\\\\h(2)= 100\\\\[/tex]
You should find that h(3) = 100 as well. I'll let you do those steps.
Since h(2) = 100 and h(3) = 100, this confirms the two solutions we found earlier.
