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Find exit speed at the bottom of the ramp using kinematics only

  • Thread starter spartan55
  • Start date
  • #1
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Homework Statement


A professional skier's initial acceleration on fresh snow is 90% of the acceleration expected on a frictionless, inclined plane, the loss being due to friction. Due to air resistance, his acceleration slowly decreases as he picks up speed. The speed record on a mountain in Oregon is 180 kilometers per hour at the bottom of a 29.0deg slope that drops 197 m. What exit speed could a skier reach in the absence of air resistance (in km/hr)? What percentage of this ideal speed is lost to air resistance?


Homework Equations


We are only on kinematics....
(v_final)^2 = (v_initial)^2 + 2*(a_parallel)*(x_final - x_initial) , where a_parallel = g*sin(29)


The Attempt at a Solution


I used trig to solve for the length of the ramp:
l*sin29 = 197
l = 406.35 m
Then I plugged this into the above kinematics equation and solved for v_final:
(v_final)^2 = 0 + 2*g*sin(29)*(406.35 - 0)
v_final = 62.14 m/s
I converted this to km/hr:
62.14 m/1s * 1km/1000m * 3600s/1hr = 223.7 km/hr, but this isn't the correct answer. I'm not sure where I went wrong.
 

Answers and Replies

  • #2
SammyS
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Homework Statement


A professional skier's initial acceleration on fresh snow is 90% of the acceleration expected on a frictionless, inclined plane, the loss being due to friction. Due to air resistance, his acceleration slowly decreases as he picks up speed. The speed record on a mountain in Oregon is 180 kilometers per hour at the bottom of a 29.0deg slope that drops 197 m. What exit speed could a skier reach in the absence of air resistance (in km/hr)? What percentage of this ideal speed is lost to air resistance?


Homework Equations


We are only on kinematics....
(v_final)^2 = (v_initial)^2 + 2*(a_parallel)*(x_final - x_initial) , where a_parallel = g*sin(29)


The Attempt at a Solution


I used trig to solve for the length of the ramp:
l*sin29 = 197
l = 406.35 m
Then I plugged this into the above kinematics equation and solved for v_final:
(v_final)^2 = 0 + 2*g*sin(29)*(406.35 - 0)
v_final = 62.14 m/s
I converted this to km/hr:
62.14 m/1s * 1km/1000m * 3600s/1hr = 223.7 km/hr, but this isn't the correct answer. I'm not sure where I went wrong.
Look at the phrase in red, especially the 90%.
 
  • #3
4
0
Ahh yes that is what I forgot. Thanks Sammy!
 

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