- #1
drummerscott
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A ski jumper starts with a horizontal take off velocity of Vo=20.0m/s and lands on a hill inclined at 30 degrees. (Note says to use a coordinate system that is not horizontal and vertical, so I rotated the coordinate system to make the ski jumper "take off" at Vo at the angle of 30 degrees. Thus x is "down" the hill and y is perpendicular)
Find: Time, distance
Here is what I have done:
(given) Ax=0 Ay=gravity Vo=20.0 theta=30
Voy=Vosin(theta)
Voy=20sin(30)
Voy=10
VFY=Voy+Ayt (solved for t)
t=(VFY-Voy)/AY
t=(0-10.0)/-9.81
t=1.02s <---Since this is only at the top of the parabola I multiplied by 2 and got 2.04s (when the skier lands)
Distance
x=xo+Voxt +1/2at2
x=0+Voxt+0
x=17.3(2.04) <-----Vox=20cos(theta)
x=35.3m
The last question is: What is the component of the ski jumpers velocity along the inclined landing hill when he lands?
The relationship between the two velocities would be the Pythagorean theorem (I think) but I'm not sure what I'm actually trying to find here.
Thanks
Find: Time, distance
Here is what I have done:
(given) Ax=0 Ay=gravity Vo=20.0 theta=30
Voy=Vosin(theta)
Voy=20sin(30)
Voy=10
VFY=Voy+Ayt (solved for t)
t=(VFY-Voy)/AY
t=(0-10.0)/-9.81
t=1.02s <---Since this is only at the top of the parabola I multiplied by 2 and got 2.04s (when the skier lands)
Distance
x=xo+Voxt +1/2at2
x=0+Voxt+0
x=17.3(2.04) <-----Vox=20cos(theta)
x=35.3m
The last question is: What is the component of the ski jumpers velocity along the inclined landing hill when he lands?
The relationship between the two velocities would be the Pythagorean theorem (I think) but I'm not sure what I'm actually trying to find here.
Thanks