This one's for you, Dave

First off there must be some ungodly spin on that star bit to come down from above and knock the Goomba UPWARDS.

However, leaving aside that bit of counter-intuition, it's pretty straightforward. Linear momentum is conserved in both relevant dimensions.

Since it looks like nifty unicode and stuff won't work here, I'll need to define variables a little more clunkily.

m = mass of bit
M = mass of Goomba
v = speed of bit (vix = initial in x, viy = initial in y, vfx = final in x, vfy = final in y)
V = speed of Goomba (same splitting).

m = 1 kg
M = 3 kg

I need to find the x and y components of the velocities given. I prefer to work with angles less than 90 and just manually put in the signs. 244 degrees equals 26 degrees down from negative x-axis.

vix = 10 m/s * cos(74) = 2.76 m/s
viy = -10 m/s * sin(74) = -9.61 m/s
Vix = Viy = 0 m/s
vfx = -2 m/s * cos(26) = -1.80 m/s
vfy = -2 m/s * sin(26) = -0.88 m/s

Okay, now I can begin.

m*vix + M*Vix = m*vfx + M*Vfx (1)
m*viy + M*Viy = m*vfy + M*Vfy (2)

Solve (1) for the desired Vfx:

Vfx = (m*vix + M*Vix - m*vfx)/M = 1.52 m/s

Solve (2) for the desired Vfy:

Vfy = (m*viy + M*Viy - m*vfy)/M = -2.91 m/s

Oops, the Goomba doesn't go up, he goes down. In fact, the star bit would have to have a LARGER speed after the impact in order for the Goomba to go up. So part C has the answer "zero seconds" because he doesn't go up in the first place.

The speed is just the square root of the sum of the squares of the components, or 3,28 m/s. He's going down and to the right, tan(angle of depression) = 2.91/1.52, so an angle of 62.4 degrees below the positive x axis. But since he's ON the ground already, at most he slides to the right at 1.52 m/s, since the ground absorbs all the vertical component.