Kinematics of jet-ski - how far does Sam land?

[habibclan]

Homework Statement

Sam (5) kg takes off up a 50-m high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis titled at 10 degrees after becoming airborne, as shown in the Figure. How far does Sam land from the base of the cliff?

I drew the picture on paint. here is the link for it:
http://i196.photobucket.com/albums/aa59/aliatehreem/Sam.jpg

Homework Equations

F= ma
vf^ 2 = vi ^2 + 2ad
vx= v* cos theta
vy= v* sin theta

yf= yo + vt+ 0.5 *a* t^2

The Attempt at a Solution

Find length of slope:

h= 50/ (sin 10)

First I have the acceleration parallel to the motion.
F= ma
ma= F (thrust) - weight parallel
ma= 200 - mg (sine theta)
a= 0.96318

Find velocity at the end of the slope.

vf^ 2 = vi ^2 + 2ad
vf^ 2 = 0 + 2(.96318)(50/ (sin 10) )
vf = 23.55 m/s

This velocity is parallel to the slope.

To find velocity after Sam gets off the slope, in the vertical and horizontal direction, the x and y component velocities are:

vx= v* cos theta= 23.55* cos10= 23.2
vy= v* sin theta= 23.55* sin 10= 4.09

Since vertical displacement is -50 m, find time, and sub into horizontal component.

yf= yo + vt+ 0.5 *a* t^2
-50= 0 + 4.09 t + 0.5 * -9.81 * t^2
t= 3.637 s

Sub into horizontal velocity to find distance:

xf= xo + vt
xf = 23.2 (3.637) = 84.4 m

The correct answer is 110 m. Can someone please tell me what I am doing wrong? Thanks in advance!

 

[ace123]

First I have the acceleration parallel to the motion.
F= ma
ma= F (thrust) - weight parallel
ma= 200 - mg (sine theta)
a= 0.96318

Check these calculations. I think this is where your mistake is.

 

[habibclan]

That was a typo. Sorry. Sam's mass is actually 75 kg.

 

[ace123]

ah. I knew something was off. Hold on

 

[rl.bhat]

ma= 200 - mg (sine theta)
a= 0.96318
Check this calculation

 

[habibclan]

ma= 200 - mg (sine theta)
a= 0.96318
Check this calculation

I'm getting the same answer:

ma= 200 - mg (sine theta)
75a= 200 - (75)(9.81) (sin 10 degrees)
a= 0.96318 m/s^2.

 

[rl.bhat]

Your answer appears to be correct.

 

[habibclan]

Your answer appears to be correct.

For the acceleration, you mean? The overall answer for some reason isn't turning out to be correct. I checked my calculations many times. In the beginning, I made a few error but then I corrected those, but the answer I'm supposed to be getting is 110 m, not 84.4 m :(. I hate textbook questions because they don't even have proper solutions, they just give the final answer without any explanation!

 

[Kurdt]

Sam is still subject to some thrust after he leaves the ramp if I'm reading the question correctly.

 

[habibclan]

Sam is still subject to some thrust after he leaves the ramp if I'm reading the question correctly.

Thanks a lot! That solves the mystery. I assumed that the force of thrust was only present while on the ramp. So now, I have an acceleration in both x and y directions and I can use a similar method to solve for the horizontal distance.

 

[Kurdt]

Well tell me that when you get the right answer. I just had a quick look.

 

[habibclan]

Well tell me that when you get the right answer. I just had a quick look.

I got the answer now! Yay! Thank you so much for guiding me! I love this forum! I'm feeling more confident now and learning more things by second :D.

 

[Kurdt]

Very good! I hope you'll be around helping other students one day.

 

[habibclan]

Very good! I hope you'll be around helping other students one day.

You never know!