10. If the astronaut from question #9 collides with a satellite and comes to a complete stop, where dic his momentum go? Explain this in terms of conservation of momentum .

I will answer in English.

Ok, we know that a ball is trowed up, and the maximum height is 21.4m

If we suppose that the initial height of the ball is h = 0 (this means that is trowed from the ground) the equations of movement will be:

for the acceleration we only have the gravitational acceleration, so we have:

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

for the velocity we can integrate over time and get:

v(t) =  (-9.8m/s^2)*t + v0

where v0 is the initial speed.

For the position we integrate over the time again, and as the initial position is 0m, here we do not have any integration constant.

p(t) = (1/2)(-9.8m/s^2)*t^2 + v0*t

a) the maximum height is reached when the velocity is equal to zero, this happens at the time:

(-9.8m/s^2)*t + v0 = 0

t = v0/(9.8m/s^2)

now we can replace this time in the equation for the position:

p = 21.4m = (-4.9m/s^2)*(v0/(9.8m/s^2))^2 + v0^2/(9.8m/s^2)

and solve it for v0, i will stop writing the units so it is easier to read:

21.4 = -4.9*(v0/9.8)^2 + v0^2/9.8

21.4 = v0^2*(0.05)

v0 = √(21.4/0.05) = 20.7

so the initial speed is 20.7 m/s.

b) the things i supposed are:

The initial position is p(0) = 0

there are no things like air resistance or wind.

c) we can take t = 3.05s and put this in the position equation:

p(3.05s) =  (1/2)(-9.8m/s^2)*(3.05s)^2 + 20.7m/s*3.05s = 17.5m

3.05 seconds after the launch, the ball will be 17.5 m above the ground.

d) we put t = 3.05s in the velocity equation:

v(3.05s) =  (-9.8m/s^2)*3.05s + 20.7m/s = -9.19m/s

which means that now the ball is falling down.