Projectile Motion (checking my work, one question)

[drummerscott]

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) A_x=0 A_y=gravity V_o=20.0 theta=30

V_oy=V_osin(theta)
V_oy=20sin(30)
V_oy=10

V_FY=V_oy+A_yt (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_oxt +1/2at²
x=0+V_oxt+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

 

[JaredJames]

For the first part, your working looks ok. However, just a note - if he lands on an incline, the downward section of the flight will be shorter than the upward section surely? So perhaps it isn't just doubling the halfway time. This would also apply to the distance as the incline would reduce the horizontal distance travelled. I'm not sure if you need to worry about this though.

So far as the last part goes, you can't use Pythagoras Theorem as the horizontal motion is separate from the vertical.

You need to use the SUVAT equations of motion. Once you know the flight time from the first section, you can use this with initial vertical velocity from peak of jump (0) and acceleration (gravity). Plug those into one of the SUVAT equations and you'll get the final velocity for the vertical component.

You then combine your final horizontal velocity (assuming no air resistance = 20m/s) with the final vertical velocity to get the landing velocity along the hill.

 

[drummerscott]

Thanks for the help, I ended up re-working the first part with a friend and found out a more precise answer. Also, thanks for clarifying the last part, I simply did not understand the question and your explanation, well, explained it :)

 

[JaredJames]

Happy to help.