



 
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  


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