# An object thrown directly upward is at a height of h feet…

## Solution:

We are given an equation h = –16(t – 3)^2 + 150, with the following information:

h = height of h feet

t = number of seconds

We need to determine the height, in feet, 2 seconds after it reaches maximum height. So we first need to determine the value of t when the object’s height h is the maximum. In other words, we need to determine the maximum value for this equation.

We first focus on “-16(t – 3)^2”. We ascertain that (t – 3)^2 is always either positive or 0. However, (t – 3)^2 is multiplied by –16, so if (t – 3)^2 is positive, then the product -16(t – 3)^2 is always negative. The best that we can do is to have that expression equal 0, which would yield the equation’s maximum value. We can obtain this value by letting t = 3.

We now know that the object reaches its maximum height at t = 3 (and the maximum height is 150 ft). However, we want the height of the object 2 seconds after it reaches the maximum height. Thus, we want the height at t = 5 since 3 + 2 = 5. Thus, we can plug in 5 for t and solve for h.

h = -16(t – 3)^2 + 150

h = -16(5 – 3)^2 + 150

h = -16(2)^2 + 150

h = -16 x 4 + 150

h = -64 + 150

h = 86

**Answer: B**