Posted by trisha on .
A man with a mass of 62 kg skis down a frictionless hill that is h = 3.3 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20-kg backpack and skis off a 2.0-m-high ledge. At what horizontal distance from the edge of the ledge does the man land?
At the bottom of the slope, he will have acquired a velocity of
V = sqrt(2gH) = 8.04 m/s,
and this will be aimed horizontally. This velocity gets lowered to V' when the picks up the backpack, since lineat momentum must be conserved.
V'*(62 + 20) = V*(62)
V' = 6.08 m/s.
Multipy that by the time t' that it takes to fall vertically 2.0 m from the ledge, and you will have the answer, X.
t' = sqrt(2*2.0/g)= 0.639 s
X = 6.08*0.639 = 3.88 m