A 0.20 kg ball is thrown straight up into the air with an initial speed of 11 m/s. Find the momentum of the ball at the following locations.

(a) at its maximum height

______ kg·m/s

(b) halfway to its maximum height

______ kg·m/s

I know that part a is supposed to be 0 but I dont really understand why. The final velocity would be 0, I know for the first part and momentum is velocity times mass and it would equal to 0. But I multiplied final velocity by mass for the 2nd part also but I did not get the right answer so I am confused. This is how I solved the 2nd part:

v = vo + at

0 = (11) + (-9.8)(t)

t= 1.122

So the time it would take to get to maximum height is 1.122 seconds.

delta y = vo t + 1/2 at^2

delta y = (11)(1.122) + .5 (-9.8)(1.122)^2

delta y = 6.1735

I divided the max height by 2 to get the height halfway to max and got 3.086.

v^2 = vo^2 + 2a (delta y)

v^2 = (11)^2 + 2 (-9.8) (3.086)

v^2 = 121 – 60.49

v^2 = 60.6144

v = 7.78 m/s

can someone please help?? Thank you!