Answer
The projectile is 750 feet above the ground $6.25$ seconds and $7.5$ seconds after being fired.
Work Step by Step
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.
