Question

1) A ball is dropped from a height of h metre above the ground and at the same time of instant another ball is projected upwards from the ground. The two balls meet when the upper ball falls through a distance h/3. Prove that the velocities of the 2 balls meet are in the ratio 2:1.

Answer 1

we know that

s = ut + (1/2)at²

now,

for ball 1

as u = 0, a = g and s = h/3

h/3 = (1/2)gt² ................................(1)

and

for ball 2

s = h - h/3 = 2h/3, a = -g

thus.

2h/3 = u₂t - (1/2)gt² ...........................................(2)

now,

by adding (1) and (2), we get

h = u₂t

or

t = h/u₂

thus,

(1) becomes

h/3 = (1/2).g.(h / u₂)²

or initial velocity of body 2 will be

u₂ = [(3/2).g.h]^1/2 .............................................(3)

now,

the velocity of the first ball when the two balls meet is given as

v₁ = u₁² + 2as = 0 + 2g(h/3)

or

v₁ = [(2/3)gh]^1/2 .................................................(4)

and

the velocity of the second ball when the two balls meet is given as

v²₂ = u₂² + 2as = (3/2).g.h - 2g(2h/3)

= (3/2)gh - (4/3)gh = (1/6)gh

or

v₂ = [(1/6)gh]^1/2 ......................................................(5)

so,

the ratio of two velocities is given as

v₁ / v₂ = [(2/3)gh]^1/2 / [(1/6)gh]^1/2

or

v₁ / v₂ = (2x6 / 3)^1/2

thus,

v₁ / v₂ = 2 / 1

hence proved