precalc question?

Question

1. The position of a ball thrown up from the top of a building 150 ft tall with an initial velocity of 20 ft/sec is given by the function s(t) = 16t² + 20t + 150. What is the maximum height of the ball?

Answer 1

-b/2a = t

this is algebra

all you're finding is the vertex of the parabola.
at2 + bt + 150
-16t2 + 20t + 150

-b/ 2 a = t = vertex
-20/[(2)(-16)] = t = 5/8

plug this value in to find s(t) or the height at the vertex (the highest point.)

-16(5/8)2 + 20(5/8) + 150 = -6.25 + 12.5 + 150

= 156.25 ft.

Answer 2

As the formula is written, the ball will go up forever. Perhaps you mean for the first coefficient to be -16. Assuming this to be the case:

Consider s, the height of the ball, to be the vertical axis and time, t the horizontal axis. Then find the vertex of the parabola.

Complete the square.

s = -16t² + 20t + 150
s - 150 = -16(t² - 20t/16)
s - 150 = -16(t² - 5t/4)
s - 150 -16(25/64) = -16(t² - 5t/4 + 25/64)
s - 156 1/4 = -16(t² - 5/8)²

The maximum height is 156 1/4 ft = 156.25 ft.