Help with simple algebra problem?

Question

An object is thrown directly upward from the ground. After t seconds its distance in ft. above the ground is s = 144t-16t^2.

a) After how many seconds is the object 128 ft. above ground?
b) Afer how many seconds does the object strike the ground?

Please show work. 10 pts. for best answer.

Answer 1

a.) s = 144t - 16t^2
128 = 144t - 16t^2
16(t^2 - 9t + 8) = 0
(t - 8)(t - 1) = 0

At, t = 8 and t = 1

b.) when it strikes the ground, s = 0,
0 = t(144 - 16t)
t = 0 and t = 9

At, t = 9 seconds

Answer 2

s = 144t - 16t^2.

(1) set s = 128
144t - 16t^2 = 128
16t^2 - 144t + 128 = 0
t' = 8 seconds and t'' = 1 second

(2)
set s = 144t - 16t^2 = 0
<==>
t(144 - 16t) = 0
<==>
t = 0, and t = 144/16 = 9 seconds

Answer 3

b) The equation is a parabola with roots 0 and 9

0 is the time the object was thrown

9 is the time the object will be back

a) To find the time it will reach 128ft just solve the equation

144t - 16t^2 = 128 or 16t^2 - 144t + 128 = 0

or simplifying t^2 - 9t + 8 = 0 which has the roots: 1 and 8

When it is going up, after 1s it will be at 128ft high and it will get the top and begin to come down... and at 8s it will be again at 128ft high... and going down.

Answer 4

s = 144t-16t^2.
128=144t-16^t^2
t^2 -9t +8= 0
(t-1)(t-8)= 0
t = 1 or 8 ball is at 128 feet
0 = 144t -16t^2
16t^2 -144t = 0t
t(16t-144) = 0
16t=144 --> t = 9 seconds when ball hits ground

Answer 5

a)
128=144t-16t^2
128/16=(144/16)t - (16/16)t^2
8+t^2-9t=9t - t^2-9t+t^2
t^2-9t+8=0
(t-8)(t-1)=0
t-8=0 t-1=0
t-8+8=0+8 t-1+1=0+1
t=8 t=1

b)
0=144t-16t^2
0=t(144-16t)
0=144-16t 0=t (this is the starting time)
0+16t=144-16t+16t
16t=144
(16/16)t=144/16
t=9