Answer:
[tex]a)\,v_b=28.209\,ms^{-1}\\\\b)\,s=108.17\,m[/tex]
Explanation:
a) According to conservation of energy
[tex]K.E_a+P.E_a=K.E_b+P.E_b\\\\\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_b +mgh_b\\\\mg(h_a-h_b)=\frac{1}{2}m(v^2_b-v^2_a)\\[/tex]
As skier is at rest initially so
[tex]v_a=0\\\frac{1}{2}mv^2_b=mg(h_a-h_b)\\\\v_b=\sqrt{2g(h_a-h_b)}\\\\v_b=\sqrt{2(9.8)(45-4.4)}\\\\v_b= 28.209 \,ms^{-1}[/tex]
B) The distance s to where she strikes the ground at C:
Horizontal distance covered:
[tex]s\,cos\,30^o=s_o+v_bt\\\\s_o=0\\\\s(0.866)=(28.20)t\\\\t=\frac{s}{32.56}--(1)\\[/tex]
vertical component of distance
[tex]s=s_o+v_ot+\frac{1}{2}at^2\\\\s\,sin\,30^o=0+0+\frac{1}{2}(9.8)t^2\\\\s=(9.8)t^2[/tex]
substituting (1) in above
[tex]s=(9.8)\frac{s^2}{1060.1536}\\\\s=108.17\,m[/tex]
