Question

A man throws A ball of mass 0.5 kg vertically upward with a velocity of 25 M / s find the initial momentum of the ball momentum of the ball at the Halfway mark of the maximum height

Answer 1

a) initial momentum of the ball :
mass=0.5kg
u=25m/s
p₁=uxm
=0.5 x 25
=12.5kg m/s
b) By V² - u² = 2as
s= V² - u²/2a
u= 25m/s
g=-10m/s²
V=0 m/s (final velocity)
(0-25)²/2x -10
=31.25m (  Max. height attained)
so velocity at mid point:
s/2=31.25/2=15.625m will be
v²= u² +2a(s/2)
=25²+2x(-10)x(15.625)
v²=31.25
v=17.67m/s
momentum at half way =P(half)= mv
                                                 =0.5x17.67
                                                 = 8.835 Kg -m/s

Answer 2

Answer:

Explanation:

Mass of the ball = 0.5kg (Given)

Upward velocity = u =25m/s (Given)

According to the equation -

p = u × m

= 0.5 × 25

= 12.5kg m/s

Maximum height = u²/2g

= (25)²/2×10

= 625/20

= 31.25 m

Thus, half of the maximum height = 31.25/2 = 15.625 m

v² = u² -2gh

v² = (25)² - 2 × 10 × (15.625)

v² = 625 - 312.5 = 312.5

v = 12.5√2 m/s

Momentum at the half of maximum height = mass × velocity at half of maximum height

= 0.5 × 12.5 √2

= 6.25√2 kgm/s