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 meters

Step-by-step explanation:

Using energy we get the velocity when the man gets to the bottom of the hill

mgh=1/2 m v^2

Then the velocity is the squareroot of two times the mass times the gravity constant = 9,598 m/s2

Using 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 V2^2

We 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 move

Y=1/2 g t^2

t=59 seconds

To get the horizontal distance we use

X = v t

X=4,85 meters.