**College physics- Free fall problem .**
An airplane is flying with a velocity of 85.0m/s at an angle of 20.0∘ above the horizontal. When the plane is a distance 111m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? You can ignore air resistance.
I know the initial velocities in the x and y direction will be 85cos20(x) and 85sin20(y).
Variables in the X Direction: (Vi=80m/s ) Vf= t= (ax=0m/s^2) (Delta X=?)
Variables in the Y Direction: Vi=29.071m/s Vf= t= ay=-9.8m/s^2 (Delta Y=111m)
With those variables, I know that I can solve for time in the y-direction and use that time to solve for the change in X. ChangeX=Vi(t)+(1/2)at^2 since acceleration in the x direction will be zero X=Vi(t). However, after I found time using variables in the Y-direction it is wrong.
Some guidance would be greatly appreciated.
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Last edited by YoungJedi; Feb 14th 2015 at 11:52 AM.
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