Question

A 55-kg skier starts from rest at the top of a ski jump, point A in Fig. 6–48, and travels down the ramp. If fric- tion and air resistance can be neglected, (a) determine her speed vB when she reaches the horizontal end of the ramp at B. (b) Determine the distance s to where she strikes the ground at C.

Answer 1

Explanation:

The first attachment contains the complete question while the second attachment is the solution.

Answer 2

Answer:

$a)\,v_b=28.209\,ms^{-1}\\\\b)\,s=108.17\,m$

Explanation:

a) According to conservation of energy

$K.E_a+P.E_a=K.E_b+P.E_b\\\\\frac{1}{2}mv^2_a+mgh_a=\frac{1}{2}mv^2_b +mgh_b\\\\mg(h_a-h_b)=\frac{1}{2}m(v^2_b-v^2_a)$

As skier is at rest initially so

$v_a=0\\\frac{1}{2}mv^2_b=mg(h_a-h_b)\\\\v_b=\sqrt{2g(h_a-h_b)}\\\\v_b=\sqrt{2(9.8)(45-4.4)}\\\\v_b= 28.209 \,ms^{-1}$

B)  The distance s to where she strikes the ground at C:

Horizontal distance covered:

$s\,cos\,30^o=s_o+v_bt\\\\s_o=0\\\\s(0.866)=(28.20)t\\\\t=\frac{s}{32.56}--(1)$

vertical component of distance

$s=s_o+v_ot+\frac{1}{2}at^2\\\\s\,sin\,30^o=0+0+\frac{1}{2}(9.8)t^2\\\\s=(9.8)t^2$

substituting (1) in above

$s=(9.8)\frac{s^2}{1060.1536}\\\\s=108.17\,m$