Answer
The projectile is in flight for approximately 2.6 seconds.
The horizontal distance covered is 62.4 feet
Work Step by Step
We can find the time the projectile is in flight:
$y = 24\sqrt{3}~t-16~t^2$
$24\sqrt{3}~t-16~t^2 = 0$
$(24\sqrt{3}-16~t)(t) = 0$
$t = 0$ or $(24\sqrt{3}-16~t) = 0$
$t = \frac{24\sqrt{3}}{16}$
$t = 2.6~seconds$
The projectile is in flight for approximately 2.6 seconds.
We can find the horizontal distance covered:
$x = 24~t$
$x = (24)(2.6)$
$x = 62.4~feet$
The horizontal distance covered is 62.4 feet
