A man throws a ball of mass 0.5 kg vertically upward with a velocity of 25 m/s. Find :

1. (a) the initial momentum of the ball

2. (b) momentum of the ball at the half way mark of the maximum height (given g = 10 m/s2 )

(a)

initial momentum = mass x initial velocity

or

p_i = m x u

here,

m = 0.5 kg

u = 25 m/s

so,

p_i = 0.5 x 25

thus,

initial momentum of ball will be

p_i = 12.5 kg.m/s

.

(b)

we know that

v² - u² = 2as

or

s = (v² - u²) / 2a

here,

final velocity (at max. height), v = 0

u = 25 m/s

a = -g = -10 m/s2

so,

s = (0 - 25²) / 2x-10

thus, maximum height attained 

s = 31.25m

so,

velocity at midpoint or s/2 = 31.25/2 = 15.625m will be

v'² = u² + 2a(s/2) 

or

v'² = 25² + 2x-10x(15.625)

or

v'² = 312.5

thus, velocity at half-way mark

v' = 17.67 m/s

so,

momentum at half-way mark

p' = m x v' = 0.5 x 17.67

thus,

p' = 8.835 kg.m/s