- #1

- 225

- 8

## Homework Statement

## Homework Equations

kinematics.

a=F/m

## The Attempt at a Solution

100cos(70)=342

100sin(70)=393.7

X=1/2(-9.8)t^2+100t; t=0. t=20.4

- Thread starter Robb
- Start date

- #1

- 225

- 8

kinematics.

a=F/m

100cos(70)=342

100sin(70)=393.7

X=1/2(-9.8)t^2+100t; t=0. t=20.4

- #2

SteamKing

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You might want to check the arithmetic above before going further in this calculation. Also, don't forget to indicate units.## Homework Statement

See attached.

## Homework Equations

Kinematic equations.

## The Attempt at a Solution

100cos(70)=342 (x component)

100sin(70)=393.7 (y component)

X=1/2(-9.8)t^2 + 100t = -4.9t^2 + 100t; t=0, t=20.4

V(final)= V(initial) + at =100 + (-9.8)(20.4)=-99.9m/s

I'm convinced t is not equal to 20.4.

- #3

CWatters

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- #4

- 225

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velocity at the top = 0

acceleration from g = -9.8m/s

I know my basic kinematics and a=f/m

- #5

gneill

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- #6

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- #7

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Not sure how this can be negative unless it is accounting for the height of the building?

- #8

gneill

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The g in this case is just the constant g without sign. You might try deriving the expression for the maximum height to see why this is.

Not sure how this can be negative unless it is accounting for the height of the building?

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