- #1

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I got part (a) to be 32.63 m?s but i dont know where to start for part b.

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- Thread starter steph2010
- Start date

- #1

- 8

- 0

I got part (a) to be 32.63 m?s but i dont know where to start for part b.

- #2

- 44

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- #3

- 8

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but arent i finding the where it would land on the y axis?

- #4

- 44

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- #5

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i am getting the answer to be 21.9264m is that right?

- #6

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Not what I got, what did you get for the time?

- #7

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t= 2.51 sec

- #8

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Hmm.

[tex] d_f = \frac{1}{2}at^2 + v_it [/tex]?

[tex] d_f = \frac{1}{2}at^2 + v_it [/tex]?

- #9

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now i have to answers for t 2.04sec and 2.5sec

- #10

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- #11

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what y and x equations?

- #12

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X: [tex] d_x = x_it [/tex]

You're doing the same thing you did to make sure the shell cleared the cliff and find the velocity for that (you didn't know time or velocity) except now you don't know time or distance.

- #13

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ok so t= 5.77 sec

- #14

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- #15

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0=-25=32.64t-4.905t^2

t= -.88s t=5.77s

t= -.88s t=5.77s

- #16

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First off, I assume the -25 is for the height of the cliff which isn't needed. We already proved that the shell would clear the cliff in part a, so start back at the cannon from ground level.

Secondly, the shell isn't shot horizontally, as your equation as it. It was shot 43º above horizontal, which affects the velocity.

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