IEsailor 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
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, when (t – 3)^2 is multiplied by –16, a negative number, the product will be negative. Thus, the best we can do is to have the expression -16(t - 3)^2 equal 0, which would yield the maximum value of –16(t – 3)^2 + 150. 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