ANSWER
400 ft
EXPLANATION
We have that the height of the arrow t seconds after it was shot is:
[tex]h(t)=-16t^2+160t[/tex]
We want to find the maximum height of the arrow. Since the arrow follows a parabolic path, the maximum height is reached when the vertical speed of the arrow is 0 - this is because the arrow slows down as it goes up, then stops for a moment and speeds back down.
The velocity of the arrow is the derivative of the height:
[tex]v=h^{\prime}=-32t+160[/tex]
To find the time we solve:
[tex]0=-32t+160[/tex]
Add 32t to both sides:
[tex]\begin{gathered} 0+32t=-32t+32t+160 \\ 32t=160 \end{gathered}[/tex]
And divide both sides by 32:
[tex]\begin{gathered} \frac{32t}{32}=\frac{160}{32} \\ t=5 \end{gathered}[/tex]
Now we replace t = 5 into the height equation:
[tex]h(5)=-16\cdot5^2+160\cdot5[/tex]
And solve:
[tex]\begin{gathered} h(5)=-16\cdot25+800 \\ h(5)=-400+800 \\ h(5)=400 \end{gathered}[/tex]
The maximum height of the arrow is 400 ft
