Solving Ski Jumper Physics: How Long is Airborne?

[chopramon]

A ski jumper travels down a slope and launches at a horizontal speed of 25m/s. The landing incline slopes off at 33 degrees.

How long is the jumper airborne?

Vx = 25m/s

horizontal motion d=Vt
Vertical Motion, d=v1t + 1/2at2, v1 = 0 as initial vertical speed is 0 off the launch

I am not sure how to approach this problem. Thanks for any help.

 

[Dick]

Ok, so your equations give you the vertical and horizontal distances (dx,dy) as a function of time from the launch point (0,0). So the jumper will land when intersecting the line with a slope corresponding to 33 degrees through (0,0). So what is dy/dx? Can you solve that expression for t?

 

[chopramon]

Ok excellent. That was my starting point as well. My major confusion was it the Vy/Vx that is equivalent to the slope, or is it the Dy/Dx that is equivalent to the slope. After this I am just setting Tan33=Dy/Dx or Vy/Vx and solve for t afterwards.

Can I ask why is it Dy/Dx and not Vy/Vx that is equal to slope?

Thanks for the help!

 

[Dick]

The jumper hits the slope when his position intersects the line. So it's the ratio of distances. If Vy/Vx equals the slope, that just means the jumper is moving parallel to the landing incline, which he probably won't be when he actually hits it.