Answer:
t = 0.293 s and 14.52 s
Step-by-step explanation:
The heights of the ball in feet is given by the equation as follows :
[tex]S = -2.7t^2+40t+6.5[/tex]
Where t is the number of seconds after the ball was thrown.
We need to find is the ball 18 ft above the moon's surface.
Put S = 18 ft
[tex]-2.7t^2+40t+6.5=18\\\\-2.7t^2+40t=18-6.5\\\\-2.7t^2+40t-11.5=0[/tex]
It is a quadratic equation. Its solution is given by :
[tex]t=\dfrac{-40+\sqrt{40^{2\ }-4\left(-2.7\right)\left(-11.5\right)}}{2\left(-2.7\right)},\dfrac{-40-\sqrt{40^{2\ }-4\left(-2.7\right)\left(-11.5\right)}}{2\left(-2.7\right)}\\\\t=0.293\ s,14.52\ s[/tex]
So, the ball is at first 0.293 s and then 14.52 s above the Moon's surface.
