A man with a mass of 65 kg skis down a frictionless hill that is 4.7 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 23 kg backpack and skis off a 1.7 m high ledge. At what horizontal distance from the edge of the ledge does the man land?

Answer: 4,85 metersStep-by-step explanation:Using energy we get the velocity when the man gets to the bottom of the hillmgh=1/2 m v^2Then the velocity is the squareroot of two times the mass times the gravity constant =9,598 m/s2Using energy again, we can get the velocity on the edge of the ledge (using the second mass, the one of the man plus the backpack)1/2 M1 V1^2=1/2 M2 V2^2We get V2=8,24 m/s2 Then we have to analyze the jump, horizontally, with constant velocity, and vertically, with constant acceleration equals to the gravity constant. To get the time we analyze the vertical moveY=1/2 g t^2t=59 secondsTo get the horizontal distance we use X= v t X=4,85 meters.