We know that:
- the height in feet t seconds after launch is given by
[tex]s=-16t^2+v_0t[/tex]
And we must find the times that the projectil will reach a height of 80ft and when v0 = 96 feet per second
To find it we need to replace s = 80 and v0 = 96 in the formula
[tex]80=-16t^2+96t[/tex]
Then, we must solve the equation for t
1. we must move all terms aside
[tex]80+16t^2-96t=0[/tex]
2. we must extract the common factor 16
[tex]16(5+t^2-6t)=0[/tex]
3. we must factor
[tex]16(t-5)(t-1)=0[/tex]
4. we must solve for t
We have two possible results
[tex]\begin{gathered} 1. \\ 16(t-5)(t-1)=0 \\ t-5=0 \\ t=5 \\ 2. \\ 16(t-5)(t-1)=0 \\ t-1=0 \\ t=1 \end{gathered}[/tex]
ANSWER:
A. 1, 5 seconds
