- #26

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- Thread starter kara
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- #26

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- #27

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- #28

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If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?

- #29

Doc Al

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- #30

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Alright, so i got 36.3 m/s for v0

- #31

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- #32

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So i know that its v will be 0 m/s at max heigh.

- #33

Doc Al

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Right... thekara said:So i know that its v will be 0 m/s at max heigh.

- #34

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- #35

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- #36

Doc Al

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That's right. You are looking for the value of x when y is maximum. Hint:kara said:

- #37

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well the max height is 53 m, and it reaches that height when v = 0.0 m/s

- #38

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i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2

53 = -1/2(9.8)t^2

- #39

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if i multiply both sides by 2 to get rid of the 1/2 i get

106 = -(9.8) t^2

106 = -(9.8) t^2

- #40

Doc Al

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You left out part of that equation; it should be:kara said:i plugged in all my values but get stuck at one point with a negative square root:

53 = -1/2(9.8)t^2

53 = v0t -1/2(9.8)t^2

Where v0 is the vertical component of initial velocity that you found earlier.

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