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gasar8
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Hi, I got homework and I need someone to check it, please. :)
http://shrani.si/f/1G/JS/1UO1SWSw/skakalnica.jpg
Ski jumper (m=70kg) goes down the hill with v0=9m/s. The hill has got R (radius of curvature = 8m). There is no friction (the hill is 1m above Earth - like on picture).
1. a) What is the velocity when the ski jumper just jumps?
My answer: Wk(begining)+Wp(begining)=Wk(end)+Wp(end)=
=(70*9^2)/2+70*10*1=(7*vfinal^2)/2+70*10*3,34 (h0+h)=
vfinal=5,84m/s
b) Find the highest point!
hmax=v0y^2/(2g)=(5,84*cos45)^2/20=0,85+h0=4,2m
c) What is the impulse of the hill to the jumper (vertical) when he is on the hill?
∫Fydt=G(end)-G(begining)
∫Fydt=0-m*v(end)y=0-70*sin45=-290Ns
2. Let the hill move without friction. (Mhill=1000kg) When jumper hits the hill (v0=9m/s) it is at rest.
a) Does the energy of the system (jumper+hill) preserve? Prove with Work of outer and inner forces.
I assume it does, but don't know how to prove it.?!
b) What is the velocity (to the observer on earth) when jumper leaves the hill? At what angle does it point?
I am not sure here, whether I use the conservation of energy equation or mumentum equation?:
m*g*h0+(m*v0^2)/2=mgh1+(m*v1^2)/2+(Mhill*v1^2)/2=
=1,5m/s angle=45°
or:
m*v0=(m+Mhill)v1=0,589m/s.
c) What is the velocity of the hill?
Is it here possible that I use here the equation of mumentum to get this vx and that in (b) I use energy conservation to get v=1,5. And get angle (in b) like cosθ=v/v1...θ=67°
Thanks for help. :)
http://shrani.si/f/1G/JS/1UO1SWSw/skakalnica.jpg
Ski jumper (m=70kg) goes down the hill with v0=9m/s. The hill has got R (radius of curvature = 8m). There is no friction (the hill is 1m above Earth - like on picture).
1. a) What is the velocity when the ski jumper just jumps?
My answer: Wk(begining)+Wp(begining)=Wk(end)+Wp(end)=
=(70*9^2)/2+70*10*1=(7*vfinal^2)/2+70*10*3,34 (h0+h)=
vfinal=5,84m/s
b) Find the highest point!
hmax=v0y^2/(2g)=(5,84*cos45)^2/20=0,85+h0=4,2m
c) What is the impulse of the hill to the jumper (vertical) when he is on the hill?
∫Fydt=G(end)-G(begining)
∫Fydt=0-m*v(end)y=0-70*sin45=-290Ns
2. Let the hill move without friction. (Mhill=1000kg) When jumper hits the hill (v0=9m/s) it is at rest.
a) Does the energy of the system (jumper+hill) preserve? Prove with Work of outer and inner forces.
I assume it does, but don't know how to prove it.?!
b) What is the velocity (to the observer on earth) when jumper leaves the hill? At what angle does it point?
I am not sure here, whether I use the conservation of energy equation or mumentum equation?:
m*g*h0+(m*v0^2)/2=mgh1+(m*v1^2)/2+(Mhill*v1^2)/2=
=1,5m/s angle=45°
or:
m*v0=(m+Mhill)v1=0,589m/s.
c) What is the velocity of the hill?
Is it here possible that I use here the equation of mumentum to get this vx and that in (b) I use energy conservation to get v=1,5. And get angle (in b) like cosθ=v/v1...θ=67°
Thanks for help. :)
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