**an arrow shot into the air is 144t-4.9t squared meters above the ground t seconds after it is released. during what period of time is the arrow above 100 meters?**

**Answer:**

f(t)= 144t-4.9t^2

100=144t-4.9t^2

4.9t^2-144t+100=0

by quadratic formula

x= (-b+sq.root(b^2-4ac))/2a

x=(144sq.root(144^2-4*4.9*100))/2*4.9

x= 28.7 seconds -> answer