Question 1209012: If an archer shoots an arrow straight upward with an initial velocity of 160 ft/sec from a height of 8 ft, then its height above the ground in feet at time t in seconds is given by the function h(t) = -16t^2 +160t + 8
1. What shape would this function make when graphed? How do you know?
2. Will this shape have an obvious minimum value, an obvious maximum value, neither, or both? How do you know?
3. Where in the graph would you find the “optimal” value (either highest or lowest value)? Find, through calculation, the t-coordinate for that “optimal” value.
4. Find, through calculation, the h(t)-coordinate for the “optimal” value?
5. Interpret the results from Step 3 and Step 4. (I.e., what do those numbers mean?)
Answer by ikleyn(52835)   (Show Source):
You can put this solution on YOUR website! .
If an archer shoots an arrow straight upward with an initial velocity of 160 ft/sec
from a height of 8 ft, then its height above the ground in feet at time t in seconds
is given by the function h(t) = -16t^2 +160t + 8
1. What shape would this function make when graphed? How do you know?
2. Will this shape have an obvious minimum value, an obvious maximum value, neither, or both? How do you know?
3. Where in the graph would you find the “optimal” value (either highest or lowest value)? Find, through calculation, the t-coordinate for that “optimal” value.
4. Find, through calculation, the h-coordinate for the “optimal” value?
5. Interpret the results from Step 3 and Step 4. (I.e., what do those numbers mean?)
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(1) The shape is a parabola. I know it because the plot of every quadratic function
is a parabola.
(2) This parabola has an obvious maximum. I know it, because the leading coefficient at t^2
is negative, and every quadratic function with the negative leading coefficient represents
a downward parabola, which has a maximum.
(3) For the general formal quadratic function f(t) = at^2 + bt + c, the "t-coordinate" of its
optimum value is t =  , which in this given case is t =  =  = 5 seconds.
(4) The h-coordinate of the optimum is h(5) = -16*5^2 + 160*5 + 8 = -400 + 800 + 8 = 408 ft.
(5) These numbers mean that the arrow will reach its maximum height of 408 ft at t= 5 seconds
after the shoot.
Solved completely.
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