Question

a ball is dropped from a height of h meter above the ground the two ball meet when the upper ball falls through a distance of h/3prove that the velocities of the two ball when they meet are in the ratio 2:1​

Answer 1

Solution:

Given:

A ball is dropped from a height of h meter above the ground the two ball meet when the upper ball falls through a distance of h/3.

We are going to use:

$\implies$ $\boxed{\sf{s = ut + (\frac{1}{2})at^{2}}}$

For ball 1:

$\bullet$ u = 0

$\bullet$ a = g

$\bullet$ s = h/3

$\implies$ $\boxed{\sf{\frac{h}{3} = (\frac{1}{2})gt^{2}}}$ - - - - (1)

For ball 2:

$\bullet$ s = h - h/3 = 2h/3

$\bullet$ a = -g

$\implies$ $\boxed{\sf{\frac{2h}{3} = u_{2}t-(\frac{1}{2})gt^{2}}}$ - - - - (2)

By adding (1) and (2), we get: $\boxed{\sf{h = u_{2}t}}$

Initial velocity of body 2 will be:

$\implies$ u2 = [(3/2).g.h]1/2 - - - - (3)

Now:

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

$\implies$ v1 = u1^2 + 2as = 0 + 2g(h/3)

$\implies$ v1 = [(2/3)gh]1/2 - - - - (4)

And,

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

$\implies$ v2^2 = u2^2 + 2as

$\implies$ v2^2 = (3/2).g.h - 2g(2h/3)

$\implies$ v2^2 = (3/2)gh - (4/3)gh

$\implies$ v2^2 = (1/6)gh

That is:

$\implies$ v2 = [(1/6)gh]1/2 - - - - (5)

So,

The ratio of two velocities is given as:

$\implies$ v1 / v2 = [(2/3)gh]1/2 / [(1/6)gh]1/2

$\implies$ v1 / v2 = (2x6 / 3)1/2

$\implies$ v1 / v2 = 2 / 1

$\implies$ v1 / v2 = 2 : 1

Hence proved.

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Answered by: Niki Swar, Goa❤️