Gmat09_5ALL 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

caspermonday wrote:

Brent Hanneson wrote:
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 (h) = 150-16(5-3)^2
=150-16(2)^2
=150-64
=86
Answer: B

Thank you for solution but I thought a little bit differently:

16(t-3)^2 is minimum when it is equal to 1 (cause the object can't fly negative amount of time).

To be equal to 1, (t-3) should equal to 1/4 (because in such case 16(t-3)^2=16(1/4)^2)

When we solve the equation t-3=1/4, we get that t=13/4. This is the time needed to reach maximum height (150-1=149)

Now we are interested in what happens in 2 seconds:
150-16(t-3+2seconds)^2=150-16(13/4-3+2)^2=69 meters

I know that there is no such answer among options but I don't see how my solution is flawed. Besides, we are never told that the number of seconds is integer.

Could anyone correct me? Thank you!