Problem Solving

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