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 and momentum of the ball at the Halfway mark of the maximum height

Answer 1

Given ,
mass of body = 0.5 Kg
initial velocity of body = 25 m/s

we know,
momentum = mass × velocity
so,initial momentum of body = 0.5 × 25 = 12.5 kg.m/s

now,
maximum height = u²/2g = (25)²/2×10 = 625/20 = 31.25 m
half of 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 half of maximum height = mass × velocity at half of maximum height = 0.5 × 12.5 √2 = 6.25 √2 kgm/s

Answer 2

Given , 
Mass of body = 0.5 Kg 
Initial velocity of body = 25 m/s
Momentum = mass × velocity 
∴Initial momentum of body = 0.5 × 25
                                           = 12.5 kg.m/s 
Again,
Maximum height = u²/2g
                          = (25)²/2×10
                          = 625/20
                          = 31.25 m 
The half would be = 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 
It is known to us that,
Momentum at half of maximum height = Mass × Velocity at half of maximum ∴Height = 0.5 × 12.5 √2
              = 6.25 √2 kgm/s