+0  
 
0
318
1
avatar

An extreme skier, starting from rest, coasts down a mountain that makes an angle of 30.9 ° with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.241. She coasts for a distance of 19.0 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.76 m below the edge. How fast is she going just before she lands?

physics
Guest Oct 20, 2014

Best Answer 

 #1
avatar+26322 
+5

1.  On the cliff.

Resolving forces parallel to the surface we have m.g.sinθ - μ.m.g.cosθ down the slope where θ is 30.9° and μ is 0.241;  m is mass and g ia gravitational acceleration.

From Newton's 2nd law of motion this must equal m.a where a is (constant) acceleration.  

This means a = g.(sinθ - μ.cosθ)

 

Hence velocity, v, at the edge of the cliff is v = √(2.a.s)  where s is 19 metres.

 

2.  Off the cliff.

The horizontal velocity, vh, is constant (ignoring air resistance etc.) at  vh = v.cosθ.

The initial vertical velocity, vv, is vv = v.sinθ.

For the vertical motion we have vvf2= vv2 + 2.g.sv  where vvf is final vertical velocity and sv is 3.76m.

 

3. Landing

The magnitude of the resultant velocity, vr, when she lands is given by:  vr = √(vvf2 + vh2

 

You can insert the numbers.

.

Alan  Oct 21, 2014
Sort: 

1+0 Answers

 #1
avatar+26322 
+5
Best Answer

1.  On the cliff.

Resolving forces parallel to the surface we have m.g.sinθ - μ.m.g.cosθ down the slope where θ is 30.9° and μ is 0.241;  m is mass and g ia gravitational acceleration.

From Newton's 2nd law of motion this must equal m.a where a is (constant) acceleration.  

This means a = g.(sinθ - μ.cosθ)

 

Hence velocity, v, at the edge of the cliff is v = √(2.a.s)  where s is 19 metres.

 

2.  Off the cliff.

The horizontal velocity, vh, is constant (ignoring air resistance etc.) at  vh = v.cosθ.

The initial vertical velocity, vv, is vv = v.sinθ.

For the vertical motion we have vvf2= vv2 + 2.g.sv  where vvf is final vertical velocity and sv is 3.76m.

 

3. Landing

The magnitude of the resultant velocity, vr, when she lands is given by:  vr = √(vvf2 + vh2

 

You can insert the numbers.

.

Alan  Oct 21, 2014

6 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details