- #1

- 30

- 0

0 = (22)t + 1/2(-9.8)t^2

I'm tryin to isolate t, so i can solve for time. I know you can factor it out as follows..

0 = [22 + 1/2(-9.8)t]t

Thats where I am stuck... how do i get this in the form.. t = xxxxx

Thanks for the help..

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- Thread starter jKotha
- Start date

- #1

- 30

- 0

0 = (22)t + 1/2(-9.8)t^2

I'm tryin to isolate t, so i can solve for time. I know you can factor it out as follows..

0 = [22 + 1/2(-9.8)t]t

Thats where I am stuck... how do i get this in the form.. t = xxxxx

Thanks for the help..

- #2

- 1,235

- 1

[tex] t = 0 [/tex] or [tex] 22-4.9t = 0, t = 4.49 [/tex]

- #3

- 30

- 0

where did the 4.9 come from? nevermind.. figured that out.. THANKS!>. i feel really dumb now because i shoulda known that.. sometimes i complicate myself too much.

Last edited:

- #4

- 135

- 0

Picking up from u

0 = [22 + 1/2(-9.8)t]t

Therefore

t = 0 and 22 + 1/2 (-9.8)t =0

finally:

t= 0, 4.49s

I think it is the second answer that you probably want. Since this type of question is probably asking when the onject will return to its original position.

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