- #1

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## Homework Statement

a) A ball is thrown in the -y direction off of a cliff with a velocity of 7m/s. If the ball takes 1.45s to reach the ground, how high off of the ground is the cliff? (Answer: -20m)

## Homework Equations

Vf=Vi+a(t)

Displacement=Vi(t)+1/2a(t)^2

Vf^2=Vi^2+2a(Displacement)

Displacement= 1/2(Vf+Vi)(t)[/B]

## The Attempt at a Solution

__Known Values:__[/B]

- Acceleration(a): -9.8m/s^2
- Time(t):1.45s
- Final Velocity(Vf): -7m/s
- Initial Velocity(Vi): 0m/s
- Displacement:?

I plug in the numbers and I'm not getting -20m as my answer. What am I doing wrong? or what am I not doing?