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An object is projected directly upward from the ground. Its distance in feet from the ground in seconds s=96t−16t2.

(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
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1+0 Answers

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 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

 

 

cool cool cool

CPhill  Oct 21, 2017
edited by CPhill  Oct 21, 2017

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