First, lets see what info we have. The proyectile formula is:
[tex]\begin{gathered} h=-16t^2+v_0t+h_0 \\ \end{gathered}[/tex]
Where v0 is the initial velocity and h0 is the initial speed.
The initial speed is 210ft per second, and since the arrow is shot from ground level, h0 = ft
Thus, the proyectile formula is:
[tex]h=-16t^2+210t[/tex]
And now we want to find the time the arrow is at 166 ft. This means we want h = 166ft
We write:
[tex]\begin{gathered} 166=-16t^2+210t \\ 0=-16t^2+210t-166 \end{gathered}[/tex]
And now we can use the quadratic equation. For a quadratic equation of the form:
[tex]0=ax^2+bx+c[/tex]
The solutions are:
[tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case:
a = -16
b = 210
c = -166
We write:
[tex]\begin{gathered} x_{12}=\frac{-210\pm\sqrt{210^2-4\cdot(-16)\cdot(-166)}}{2\cdot(-16)}=\frac{-210\pm\sqrt{44100-10624}}{-32}=\frac{210\pm182.96}{32} \\ x_1=\frac{210+182.96}{32}=12.28 \\ x_2=\frac{210-182.96}{32}=0.84 \end{gathered}[/tex]
Thus the arrow goes up to 166ft after 0.84 seconds, and then once again (in it's path down) at 12.28 seconds
