Question

A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s​ (in feet) of the ball from the ground after t seconds is s=96t-16t^2.

​(a) At what time t will the ball strike the​ ground?
​(b) For what time t is the ball more than 128 feet above the​ ground?

Answer 1

Step-by-step explanation:

a)

when is s = 0 ?

0 = t(96 - 16 t)

t = 0 of course, it starts at 0

t = 96/16 = 6 seconds

b)

when is s = 128 (two times, on the way up and on the way down)

128 = 96 t - 16 t^2

8 = 6 t - t^2

t^2 - 6 t + 8 = 0

(t-4)(t-2) = 0

so between 2 seconds and 4 seconds