an object of mass 0.5kg at rest, started to fall aheigh of 180cm from the surface of earth. it's momentum when it reaches the earth's surface is

m=0.5kg

h = 180 cm = 1.8 mh=180cm=1.8m

Initial potential energy of the object is:

E_p=m*g*hE

p

= m∗g∗h

Kinetic energy at the surface:

E_k=/frac{mv^2}{2}E

k

=

2

mv

2

According to the law of conservation of energy (assuming no air resistance):

E_p = E_kE

p

= E

k

mgh=/frac{mv^2}{2}mgh=

2

mv

2

Solving for v:

v=/sqrt{2gh}v=

2gh



p=mvp=mv

So,

p = m*v = m/sqrt{2gh}p=m∗v=m

2gh



Calculating:

p = 0.5/sqrt{2*9.8*1.8}/approx 2.97 / frac{kg*m}{s}p=0.5

2∗9.8∗1.8

≈2.97

s

kg∗m

Answer:

p / approx 2.97 / frac{kg*m}{s}p≈2.97

s

kg∗m