At what point does she lose contact with the snowball and fly off at a tangent? That is

α ≥ 48.2°

Step-by-step explanation:

The complete question is given as follows:

" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight  down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the  instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to  the skier make with the vertical?"

- The figure is also attached.

Solution:

- The skier has a mass (m) and the snowball’s radius (r).

- Choose the center of the snowball to be the zero of gravitational  potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and  departs from the snowball.

- If we ignore snow-­ski friction along with air resistance, then the one work producing force in this problem, gravity,  is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical  energy at the top of the snowball.

- Hence, From conservation of energy we have:

                       KE (α) + PE(α) = KE(α = 0) + PE(α = 0)

                       0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r

                       0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r

                        m*v(α)^2 / r = 2*m*g( 1 - cos(α) )

- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity  can hold her to that path, i.e. as long as:

                         m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )

- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal  force—or else the skier will sink into the snowball.

- The expression for α_lift becomes:

                            3*cos(α) ≥ 2

                            α ≥ arc cos ( 2/3) ≥ 48.2°

⊂⊃⊂⊃ω

step-by-step explanation:

i hope this : )

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: