- #1

drummerscott

- 4

- 0

Find: Time, distance

Here is what I have done:

(given) A

_{x}=0 A

_{y}=gravity V

_{o}=20.0 theta=30

V

_{oy}=V

_{o}sin(theta)

V

_{oy}=20sin(30)

V

_{oy}=10

V

_{FY}=V

_{oy}+A

_{y}t (solved for t)

t=(V

_{FY}-V

_{oy})/A

_{Y}

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=x

_{o}+V

_{ox}t +1/2at

^{2}

x=0+V

_{ox}t+0

x=17.3(2.04) <-----V

_{ox}=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