vittovangind wrote:An object thrown directly upward is at a height of h feet after t seconds; where h = -16(t-3)² + 150. At what height, in feet, is the object 2 seconds after it reaches its maximum height?
a) 6
b) 86
c) 134
d) 150
e) 166
The formula
h = -16 (t - 3)² + 150 allows us to determine the height of the object at any time. For what value of t is -16(t-3)² + 150 maximized (in other words, the object is at its maximum height)?
It might be easier to answer this question if we rewrite the formula as h = 150 - 16(t-3)²
To MAXIMIZE the value of h, we need to MINIMIZE the value of 16(t-3)² and this means minimizing the value of (t-3)²
As you can see,(t-3)² is minimized when
t = 3.
We want to know the height
2 seconds AFTER the object's height is maximized, so we want to know that height at
5 seconds (
3+
2)
At t =
5, the height = 150 - 16(
5 - 3)²
= 150 - 16(2)²
= 150 - 64
= 86
Answer:
B
Cheers,
Brent
