- #1

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i have no clue where do i start with.

plz help me out

sincerly

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- #1

- 47

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i have no clue where do i start with.

plz help me out

sincerly

- #2

Kurdt

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What equations do you think you need to solve this problem?

- #3

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i try problem 2 time,

my answer is different from book but they doesn't show me how to there

- #4

Kurdt

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- #5

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v^2=V^2+2aX

v^2=0+2(9.8)(1000-200)

i got V= 125

did i do anything wrong?

- #6

Kurdt

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- #7

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a=50/80=.625m/s^2

v^2=V^2+2aX v^2=0+2(.625)800=31.6m/s this is the final velocity of his 1st part where he open the parachute. correct me if i wrong

- #8

Kurdt

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The acceleration he experiences will be 9.81 - 0.625 ms^{-2} ok?

- #9

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why is that?

i did like u say

v^2=V^2+2aX v^2=0+2(9.81-.625)800

v=121.2m/s

- #10

Kurdt

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- #11

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v^2=121+2(35.3)200

v=119m/s

according to the book answer they have 24.5m/s

what did i do wrong?

- #12

Kurdt

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- #13

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if i use negative for acceleration then i got the square root of negative number for velocity

- #14

Kurdt

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No you don't because (121.2)^{2} is bigger than 400 x (-35.2).

- #15

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im sorry i forgot to square the initial velocity. I feel bad now

- #16

Kurdt

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im sorry i forgot to square the initial velocity. I feel bad now

Its fine I've done it myself.

- #17

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5^2=v^2+2(-35.2)x

i need to find initial Velocity to calculate x, im stuck again

give me hint plz

- #18

Kurdt

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You also know [itex] x_1 = 1000 - h [/itex] and [itex]x_2 = h-0[/itex] and thus you can substitute into the equation you have given to get an equation of only one variable which you can solve.

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