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!
