Answer
The projectile is in flight for approximately 3.4 seconds.
The horizontal distance covered is 357.5 feet.
Work Step by Step
We can find the time the projectile is in flight:
$y = 53.57~t-16~t^2+3.2$
$53.57~t-16~t^2+3.2 = 0$
$16~t^2-53.57~t-3.2 = 0$
We can use the quadratic formula to find the time $t$ the projectile is in flight:
$t = \frac{-(-53.57)\pm \sqrt{(-53.57)^2-4(16)(-3.2)}}{(2)(16)}$
$t = \frac{53.57\pm 55.45}{32}$
$t = -0.059, 3.4$
Since time of flight must be positive, $t = 3.4~seconds$.
The projectile is in flight for approximately 3.4 seconds.
We can find the horizontal distance covered:
$x = 105.14~t$
$x = (105.14)(3.4)$
$x = 357.5~feet$
The horizontal distance covered is 357.5 feet.
