Answer:
After 10 seconds
Step-by-step explanation:
In this problem, the height of the object after t seconds is described by the function
[tex]h(t)=-16t^2+144t+160[/tex]
where
160 ft is the initial height of the object at t = 0
+144 ft/s is the initial velocity
[tex]-32 ft/s^2[/tex] is the acceleration due to gravity (downward)
Here we want to find the time at which the object hits the ground, so the time t at which
[tex]h(t)=0[/tex]
Therefore we can write
[tex]-16t^2+144t+160 =0[/tex]
Simplifying (dividing each term by 16), we get
[tex]-t^2+9t+10=0[/tex]
This is a second-order equation in the form
[tex]ax^2+bx+c=0[/tex]
Which has solutions given by the formula
[tex]t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex] (2)
Here we have:
a = -1
b = 9
c = 10
Substituting into (2) we find the solutions:
[tex]t_{1,2}=\frac{-9\pm \sqrt{9^2-4(-1)(10)}}{2(-1)}=\frac{-9 \pm 11}{-2}[/tex]
Which gives:
[tex]t_1=10 s\\t_2 =-1 s[/tex]
Since time cannot be negative, the only solution is
t = 10 seconds
