If a rock falls for 5 seconds near the surface of the earth and with no air friction, it will reach a velocity of

-49m/s.

Explanation:

As the rock only falls, we will assume that the initial vertical velocity is zero.

We neglect the air friction, so the only force acting on the rock is the gravitational force, this means that the acceleration is -g = -9.8m/s^2.

Then we can write:

a(t) = -9.8m/s^2

To write the velocity of the rock, we must ingrate over time and get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity, and as we said above, v0 = 0m/s

Then the vertical velocity as a function of time is:

v(t) = (-9.8m/s^2)*t

Now, the question is:

"...If a rock falls for 5 seconds near the surface of the earth and with no air friction, it will reach a velocity of..."

Then we need to evaluate the velocity equation in t = 5 seconds.

v(5s) = (-9.8m/s^2)*5s = -49m/s.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//