Ski jumper leaving slope horizontally at 25 m/s on 35° incline

[flower76]

I think I have the first part of the problem but am not sure on the second part, I'm probably thinking too hard - the simplest answer is probably the correct one, but I just want to make sure.

Problem:
A ski jumper travels down in a slope and leaves the ski track moving in the horizontal direction with a speed of 25m/s. The landing incline below her falls off with a slope of 35 degrees.
a)Where does the ski jumper land on the incline?
b)How long is the ski jumper airborne?

I worked out the x and y components of the distance fallen, and found an x component of landing 89.3m downrange, which is 109m distance along the landing slope.

 

[Andrew Mason]

I think I have the first part of the problem but am not sure on the second part, I'm probably thinking too hard - the simplest answer is probably the correct one, but I just want to make sure.

Problem:
A ski jumper travels down in a slope and leaves the ski track moving in the horizontal direction with a speed of 25m/s. The landing incline below her falls off with a slope of 35 degrees.
a)Where does the ski jumper land on the incline?
b)How long is the ski jumper airborne?

I worked out the x and y components of the distance fallen, and found an x component of landing 89.3m downrange, which is 109m distance along the landing slope.

Use $h = \frac{1}{2}gt^2$ to find the time, where h is the vertical distance from the point of take-off to the point of landing.

Here is an interesting question: what is the shape of the curve that would result in the skier reaching the takeoff point in the minimum time? (Don't work it out - it is rather complicated. Try finding it on the internet).

AM

 

[flower76]

Ski Jump Physics, please check answer

I think I finally figured it out. I found the height using v0sin35 = 14.34m
Then using h=1/2gt^2 I determined that the time airborne was 2.93 seconds.

Could anyone please verify that I have done this correctly?

Thanks

 

[Andrew Mason]

I think I finally figured it out. I found the height using v0sin35 = 14.34m
Then using h=1/2gt^2 I determined that the time airborne was 2.93 seconds.

Could anyone please verify that I have done this correctly?

In my response, I had assumed that you already had worked out the horizontal distance in part a). But I see that you have to find the time first. I get a little longer time than 2.93 sec.:

$$\frac{d_y}{d_x} = -tan(35)$$

$$d_x = v_xt$$ and

$$d_y = -\frac{1}{2}gt^2$$

So: $$\frac{d_y}{d_x} = -\frac{gt^2}{2v_xt} = -tan(35)$$

$$t = \frac{2tan(35)v_x}{g} = 2*.700*25/9.8 = 3.57 sec$$

$$d_y = -.5 * 9.8 * (3.57)^2 = -62.45 m.$$

And to check:

$$\frac{d_y}{d_x} = -62.45/25*3.57 = -.70 = -tan(35)$$

AM

 

[flower76]

Thanks I think I've got it now, but I'm starting to wonder if I did the first part correctly! I will have to look at it again.

 

[kjohnson]

I believe that to set up this problem the easiest you would use a rotated coordinate system that lies along the slope instead of one that lies along the horizontal. If you use a non-rotated system like described above then you can't simply just use h=(1/2)gt^2 because you don't know what your fall height is. If you use a rotated system then you can find your x and y accelerations. Then integrate two times to get your x(t) and y(t) parametric functions. Then set y(t)=0 and you will get the time value. Last plug that value into your x(t) to find how far down the slope the skier landed. I did some quick calculations and got the follwoing:
t=5.1s
x=208.8m

 

[kjohnson]

Just did a check and it turns out i had an error in one of my angles. My results now match Andrews with a time of 3.57s. I got my x value to be 109m.

 

[zerocoll20]

i got this problem, and i can't solve too. O saw the solution but i forgot :X

 

[zerocoll20]

omg is 62m the book is wrong:P

 

[NascentOxygen]

3.57secs, 109m down the slope