maihuna 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
Answer:

Let's think about our equation:

h = -16 (t - 3)2 + 150

(I'm assuming that the 2 after the brackets means that we really have (t-3)^2, not (t-3)*2.)

Since (t-3)^2 will always be non-negative, and since -16 is negative, the first term, -16*(t-3)^2, will always be non-positive.

So, in order to maximize height, we want that first term to be 0, which will occur when (t-3)^2 = 0, which will occur when t=3.

Therefore, we reach our max height of 150 when t=3.

We want to know our height 2 seconds after that, so let's let t=5 and solve!

h = -16*(t-3)^2 + 150

h = -16*(5-3)^2 + 150

h = -16*(2)^2 + 150

h = -16*4 + 150

h = -64 + 150

h = 86... choose B.

Thanks Stuart yes it is ^2