aman88 wrote:Q: 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

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