An object is projected directly upward from the ground. Its distance in feet from the ground in seconds s=96t−16t^{2}.

(a) After how many seconds will the object be 128 feet from the ground? (Hint: Look for a common factor before solving the equation.)

(b) When does the object strike the ground?

(a) Select the correct answer below and fill in the answer box(es) to complete your choice.

A. The object will be 128 feet from the ground only after ____ second(s).

B. The object will be 128 feet from the ground after ____ second(s) and after ___ second(s)

ladiikeiii Oct 21, 2017

#1**+2 **

s = 96t−16t^2

For the first part... let s = 128 ......so we have.....

128 = 96t - 16t^2 divide through by 16

8 = 6t - t^2 rearrange as

t^2 - 6t + 8 = 0 factor

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

So

t - 4 = 0 or t - 2 = 0

add 4 to both sides add 2 to both sides

t = 4 t = 2

So.....it will be 128 ft above the ground after 2 seconds and after 4 seconds ...it is falling at the 4 second mark

For part (b), when it hits the ground, s = 0

So....see if you can solve this

0 = 96t - 16t^2

Let me give you a hint......if it takes 2 seconds to reach a height of 128 ft....it will also take 2 more seconds to reach the ground after reaching the 128 ft mark for the * second* time

CPhill Oct 21, 2017