How Far Does a Skateboarder Land from a Ramp?

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SUMMARY

The skateboarder, starting from a 1.0 m high ramp inclined at 30 degrees with an initial speed of 7.0 m/s, lands 5.65 meters from the end of the ramp. The calculations involve resolving the initial velocity into horizontal and vertical components, determining the time of flight using kinematic equations, and applying projectile motion principles. The final horizontal distance is derived from the horizontal velocity and time of flight, confirming the skateboarder's landing position.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic trigonometry for resolving vectors
  • Knowledge of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Study advanced projectile motion scenarios
  • Learn about energy conservation in motion
  • Explore the effects of friction on motion
  • Investigate real-world applications of kinematics in sports
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Physics students, educators, and anyone interested in understanding the mechanics of motion in sports, particularly skateboarding dynamics.

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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?
 

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