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2D Kinematics - Projectile Motion

  • Thread starter sirh3nry
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
Take your horizontal equation Dx = V_0x * t and put in the distance: 1.5 = V_0x * t [1]
Take your vertical equation V_fy = V_0y + a_y * t and realize the final vertical velocity is the same as the initial (down instead of up): -V_0y = V_0y - g*t [2]
Two equations, but three unknowns t, V_0x and V_0y.

I would put your last two equations together to cancel V_0 and get a relationship between V_0x and V_0y for the third equation. An alternative would be to use those two equations to replace the V_0x and V_0y in [1] and [2], cutting the variables down to just t and V_0.
 

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