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## Homework Statement

A koala throws a eucalyptus branch. It goes 1.5 meters. If it left at 43 degrees, what velocity did the branch have when it left the koala's paw?

The answer must be in vector form (m/s @ degrees).

Needed: initial velocity

Known:

x-displacement = 1.5m

x-acceleration = 0

y-displacement = 0

y-acceleration = -9.8 m/s2

angle theta = 43 degrees

## Homework Equations

Dx = V_0x * t

Dy = v_0y * t + 1/2 * a_y * t^2

V_fy = V_0y + a_y * t

V_fy^2 = V_0y^2 + 2 * a_y * Dy

V_0x = V_0 * cos(theta)

V_0y = V_0 * sin(theta)

## The Attempt at a Solution

Well, I realized that v_0y = -v_fy, and V_0x = V_fx.

Also, at the midpoint, the time is half of the total time, and Dx is 1/2*1.5.

At the midpoint V_fy = 0.

I couldn't figure out where to go from here. I figured the first plan of action was to find the time, then to find the V_0