Answer:
6.14m
Explanation:
As the man skis from top to the bottom of the hill, assuming that there's no friction, his potential energy would be converted to kinetic energy:
By law of energy conservation:
[tex] E_k = E_p [/tex]
[tex] \frac{mv^2}{2} = mgy[/tex]
[tex] v^2 = 2gy [/tex]
[tex]v = \sqrt{2gy}[/tex]
Let's g = 10 m/s2 and y = 5m. Then his speed when he reaches the horizontal section is
[tex]v = \sqrt{2*10*5} = 10m/s [/tex]
When he grabs the 2kg backpack, momentum conservation dictates is speed after
[tex]mv = Mv_2[/tex]
where m = 65 kg is the man mass before the grabbing. M = 65 + 2 = 67 kg is the total mass of the man and the bag after grabbing.
[tex]v_2 = v(m/M) = 10(65/67) = 9.7 m/s [/tex]
When he drops from a 2 m ledge, suppose he was skiing perfect horizontally before, then his initial vertical speed would be 0. Gravity g = 10 m/s is the only vertical acceleration that takes him down 2m. Since we have
[tex]s = \frac{gt^2}{2}[/tex]
where s = 2 m is the distance covered by gravitational acceleration g = 10m/s. Then the time it takes is
[tex]t^2 = \frac{2s}{g} = \frac{4}{10} = 0.4[/tex]
[tex]t = \sqrt{0.4} = 0.632 s[/tex]
This is also the time it takes for him to travel horizontally across with speed of 9.7 m/s
[tex]s_h = v_2t = 9.7*0.632 = 6.14m[/tex]
