Carcass 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. 214
Kudos for the right solution and explanation
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
_________________
Brent Hanneson - founder of Greenlight Test Prep