How Far Does a Skateboarder Land from a Ramp?

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A skateboarder on a 1.0 m high, 30-degree ramp starts with a speed of 7.0 m/s and rolls without friction. Initial calculations estimated the landing distance at 1.0016 meters from the ramp's end, but further analysis revealed a mistake. By correctly resolving the velocity components and applying kinematic equations, the skateboarder lands approximately 5.65 meters away from the ramp. The revised calculations account for both horizontal and vertical motion, leading to a more accurate result. The final distance indicates the importance of correctly applying physics principles in projectile motion scenarios.
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Homework Statement



A skateboarder starts up a 1.0 m high 30 degree ramp at a speed of 7.0 m/s. The skateboard wheels roll without friction. How far from the end of the ramp does the skateboarder touch down?

The Attempt at a Solution



ramp length=
h=1sin30=1/2
length=2m

v^2=vi^2+2ad
v^2=7^2+2(-9.8)(2)
vf=3.13m/s

vf=vi+a(Dt)
0=3.13+-9.8(t)
delta t=.32s

Xf=xi+vi(dT)
x=0+3.13(.32)
xf=1.0016

yf=yi+vi(dt)+1/2a*(dt)^2
yf=1+0(.32)+1/2(-9.8)(.32)^2
yf=0

The skateboarder lands 1.0016 meters from the end of the ramp.




I know I'm missing something somewhere... can someone please help me find it.
 
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Hmm... That wasn't even close... Let me try this again.

cos30(7)=vx= 6.06 m/s
sin30(7)=vy= 3.5 m/s

-1=3.5(t)+.5(-9.8)(t^2)
0=-4.9t^2+3.5t+1
t= 0.933 s

x= (6.06)(.933)+.5(0)(.933)^2
x= 5.65 m

How does that look?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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