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?
+33614 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.
.
+33614 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.
.
