We are given a function which models the height in feet of a projectile.
[tex]s=-16t^2+v_0t[/tex]
Given that the initial velocity;
[tex]v_0=96\text{ ft per second}[/tex]
We can determine what time the projectile would return to the ground by finding the value of the equation when;
[tex]s=0[/tex]
This is because, the heigh at the ground is zero foot.
Hence we would have the function as;
[tex]-16t^2+96t=0[/tex]
We begin by factoring both sides;
[tex]\begin{gathered} -16t^2+96t=0 \\ 16t(-t+6)=0 \end{gathered}[/tex]
Refine the parenthesis and re-write as;
[tex]16t(6-t)=0[/tex]
We shall now apply the rule which states;
[tex]\begin{gathered} \text{If;} \\ ab=0 \\ \text{Then;} \\ a=0,b=0 \end{gathered}[/tex]
Therefore, we would have;
[tex]\begin{gathered} 16t(6-t)=0 \\ 16t=0 \\ t=\frac{0}{16} \\ t=0 \\ \text{Also;} \\ 6-t=0 \\ 6=t \end{gathered}[/tex]
This shows that the projectile would reach the ground after 6 seconds. The other result t = 0 shows that the projectile was on the the ground at 0 seconds (just before take-off).
ANSWER:
[tex]t=0,6[/tex]
