Finding Momentum of 0.12 kg Ball Thrown Upward at 20 m/s

[missashley]

Homework Statement

A 0.12 kg ball of dough is thrown straight up into the air with an initial speed of 20 m/s.
Acceleration of gravity is 9.81 m/s^2

What is its momentum halfway to its maximum height on the way up? Answer in kg*m/s
(Don't worry about sigfigs)

Homework Equations

hmax= Vi^2 / 2g
Vf^2 = Vi^2 + 2ad

The Attempt at a Solution

hmax = (20)^2 / 2(9.81)
hmax = 400 / 19.62
hmax = 20.38735984 m

Since we are finding the halfway momentum you divide that by 2
20.38735984 / 2 = 10.19367992 m

Vf^2 = Vi^2 + 2ad
Vf^2 = (20)^2 + 2(-9.81)(10.19367992)
Vf^2 = 400 + - 200
Vf^2 = 200
Vf = square root of 200 = 14.14213562 m/s

p = mv
p = (0.12)(14.14213562)
p = 1.697056275 kg*m/s

Did I do this right?

 

[katchum]

Yep, now try to find out how you got those two formula?