Hi Heather,

I think your function should be

\[H(t) = -16 t^2 + 240 t + 4000.\]

The instructions say that $H(t)$ is the height of the arrow above the canyon floor, in feet, $t$ seconds after being shot, $t>0.$ This expression is also valid when $t = 0.$ Let's try it. In the expression for $H(t)$ substitute the number $0$ for $t,$ each time it appears. Thus

\[H(0) = -16 \times 0^2 + 240 \times 0 + 4000\]

but $0^2 = 0, -16 \times 0 = 0$ and $240 \times 0 = 0$ and hence

\[H(0) = -0 + 0 + 4000 = 4000 \mbox{ feet.}\]

$t = 0$ is the instant when the arrow is shot and at that time we already knew than the arrow was $4000$ feet above the floor of the canyon so $H(0) = 4000$ feet agrees with the information given.

Where is the arrow $2$ seconds after it is shot? In other words what is $H(2)?$

\[H(2) = -16 \times 2^2 + 240 \times 2+ 4000.\]

Evaluate this expression to find the height of the arrow above canyon floor 2 seconds after it is shot.

Where is the arrow $2$ seconds after it is shot?

Penny