Aman 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?

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.

12*c

step-by-step explanation:

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.