## Precalculus (6th Edition)

The projectile is 750 feet above the ground $6.25$ seconds and $7.5$ seconds after being fired.
To find the time at which the projectile is 750 feet above the ground, substitute 750 to $s$ then solve for $t$ to have: $$s=-16t^2+220t \\750=-16t^2+220t \\16t^2-220t+750=0$$ The equation above has $a=16, b=-220,$ and $c=750$. Solve the equation using the quadratic formula to obtain: $x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\x=\dfrac{-(-220) \pm \sqrt{(-220)^2-4(16)(750)}}{2(16)} \\x=\dfrac{220\pm \sqrt{48,400-48,000}}{32} \\x=\dfrac{220 \pm \sqrt{400}}{32} \\x=\dfrac{220 \pm 20}{32} \\x=\dfrac{220}{32}\pm \dfrac{20}{32} \\x=\dfrac{55}{8} \pm \dfrac{5}{8} \\x_1=\dfrac{55}{8}-\dfrac{5}{8}=\dfrac{50}{8}=\dfrac{25}{4}=6.25 \\x_2=\dfrac{55}{8} + \dfrac{5}{8}=\dfrac{60}{8}=\dfrac{15}{2}=7.5$ Therefore, the projectile is 750 feet above the ground after 6.25 seconds and after 7.5 seconds.