Need A Brilliant Mind! Momentum Question

A .12 kg ball of dough is thrown straight up into the air with an initial speed of 20 m/s. The acceleration of gravity is 9.81 m/s^2. What is the momentum half way up to it's maximum point? Answer in units of kg*m/s?

Please help

Momentum = mass * velocity, you have the mass so let's find the velocity

u = initial velocity = 20 m/s
v = final velocity
a = -9.81 m/s² (negative because it is downward)
t = time
s = distance traveled

v² = u² + 2as

When the ball reaches it's high point, it's velocity is 0 (right before it starts falling), so plug everything in & solve for s to figure out how high it goes

0² = 20² + 2(-9.81)s

0 = 400 - 19.62s

s = (-400)/(-19.62) = 20.387 meters, this is how high the ball goes.

Now use the same equation with s = to half that & solve for velocity

v² = 20² + 2(-9.81)(20.387/2) = 400 - 200 = 200

v = √200 = 14.14 m/s

So momentum = 0.12kg * 14.14 m/s = 1.7 kg*m/s

I believe the mass of the dough ball = 0.12 kg
{always a good idea to put leading 0 before a decimal dot, which can be missed}
time after release when ball reaches its max height = 20/9.8 = 2.04082 s = t
the max height of ball = 1/2gt² = (0.5)(9.8)(2.04082)² = 20.4 m
half way up = 10.2 m
speed V at half way up = speed V when falling from max height a distance = 10.2 m:
V = √2gh = √(2)(9.8)(10.2) = √199.92 = 14 m/s
mV = (0.12)(14) = 1.7 kgm/s ANS

At maximum height final velocity ,v = 0
v^2 = u^2 + 2as
0 = 20^2 - 2 x 9.81 x s
2 x 9.81 x s = 400
9.81s = 200
s = 200/9.81
So halfway point is at distance s = 100/9.81
v^2 = 20^2 - 2 x 9.81 x 100/9.81 = 400 - 200 = 200
v = root 200 = 14.14m/s
momentum = mv = 0.12kg x 14.14 = 1.7 kgm/s