physics help....velocity?
A skier starts from rest at the top of a 50 m hill, skis down a 36° incline into a valley, and continues up a 30 m high hill. Both hill heights are measured from the valley floor. Assume that you can neglect friction and the effect of ski poles.
(a) How fast is the skier moving at the bottom of the valley?
______m/s
(b) What is the skier's speed at the top of the next hill?
______m/s
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For this question let's use conservation of energy.
The potential energy (U) that the skier has at the top of the hill is equal to the kinetic energy (k) at the bottom.
a.) U = m*g*h where m= mass, g= acceleration due to gravity (9.81 m/s^2), h = height of hill
k = 1/2 * m*v^2 where m = mass, v= velocity (m/s)
If we set the energys equal we can use the expression:
mgh = 1/2 mv^2, Since we have m on both sides it cancels out
gh= 1/2 v^2-----> Divide both sides by 1/2 and take the square root of both sides.
sqrt(2*g*h) = v
v=sqrt(2*9.81*50) -----> v= 31.32 m/s
b.) We can use the same thing but we will take the difference between the two hill heights.
50-30 = 20
v=sqrt(2*9.81*20)
v= 19.8 m/s