# Frictionless Ski Jump at an angle

1. Sep 25, 2008

### Phoenixtears

1. The problem statement, all variables and given/known data
Sam (65 kg) takes off (from rest) up a 50 m high, 10° frictionless slope on his jet-powered skis. The skis have a thrust of 260 N. He keeps his skis tilted at 10° after becoming airborne, as shown in Figure P6.43. (Figure attachment not working- it's basically just drawing what is described)

It is worth repeating that the slope offers no friction to the skis.

What is his acceleration on the incline?
2.30 m/s^2

What is his velocity at the top of the incline?
m/s at 10 degrees above the horizontal

How far does Sam land from the base of the cliff?
m

2. Relevant equations

M=Fa
Max= F1- F2
May= F1-f2

3. The attempt at a solution
I began by drawing a force diagram (I tilted the axes 10 degrees). Then wrote general equations: Max= Thrust- mg(sinx) and May= N - mg(cosx). Putting in the actually numbers, and then using the F=ma equation I got that acceleration is 2.30 m/s2. The 10 degress for B) was practially a given.

Now, to find the velocity I set up a chart, horizontal and verticle. For verticle I was able to use the three motion equations and got this:
Verticle:
Delta-x= 50
V0= 0
Vf= 15.157
a= 2.30
t= 6.59

Now, all I need, I believe, is the horizantal final velocity and then I can use a^2 + b^2= c^2 to get the hypotenuse of the velocity at the edge of the incline. However, I can't develop a way to get that. For horizontal I discovered that the acceleration is 0 and the time is equal to that of the verticle: 6.59. Is there even a way to get the horizontal final velocity??? Or perhaps I'm doing this the long way?

Also, for part C, I believe that once I have the velocity, part C will simplify, for I can just find the time over the entire jump (set verticle final to 0 to find halfway time, then double it- time equals each other). Then I'd have enough information to find the total distance from start to finish. All I'd have to do is subtract the delta-x from the horizontal in the first 12 seconds. But I can only find that if I have the velocity. How can I find it?

~Phoenix

2. Sep 25, 2008

### alphysicist

Hi Phoenixtears,

I might be misunderstanding you, but this does not look right to me. You said this was for the vertical direction, and so the vertical displacement is 50m. However, the acceleration is not 2.3m/s2 in the vertical direction--it's 2.3m/s2 up (along) the incline.

For the motion on the incline, it's one dimensional motion (along a straight line), so you don't need to break it up into horizontal and vertical components.

Since you know the total velocity at the top of the incline is 10 degrees above the horizontal, as soon as you find one of the sides of the velocity triangle you can find the others using trigonometry.

(However, the acceleration is not zero in the horizontal direction.)

I think a picture would really help here. Are you looking for the horizontal distance from his starting point (at the beginning of the incline) or his launch point (at the end of the incline)?

Also, it does not sound to me like he will be landing at the same height that he left the incline at. (He hits the ground 50m below the top of the incline, right?) If so, then you cannot just double the time to reach the highest point, because the way down (from highest point to ground) is a longer path than the way up (from top of incline to highest point).