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 – Creator of greenlighttestprep.com

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