Basketball and tennis ball: Part I

For this case:

• Let $v_1$ and $v_2$ the final velocities of the basketball ball and tennis ball respectively. Same convention for masses $m_1$ and $m_2$ .

• Air resistance is negligible

• The origin for the frame of reference is height $h$

Since the balls fall from rest, the time required to fall to a height $h$ is

$\frac { 1 }{ 2 } g{ t }^{ 2 }=h\quad \quad \therefore \quad \quad t=\sqrt { \frac { 2h }{ g } }$

Therefore the velocity for both balls is

$v=gt=g\sqrt { \frac { 2h }{ g } }=\sqrt { 2gh}$

When the basketball ball reaches the ground, both balls have a velocity $-v$ , when the same ball ( the basketball one) bounces off the ground, for an instant, the basketball ball goes with a velocity $v$ meanwhile the tennis ball goes with a velocity $-v$ . As the collision is elastic, momentum and kinetic energy are conserved, this gives us two equations

${ m }_{ 1 }v-{ m }_{ 2 }v={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }v_{ 2 }.................(1)$

${ m }_{ 1 }{ v }^{ 2 }+{ m }_{ 2 }{ v }^{ 2 }={ m }_{ 1 }{ v }_{ 1 }^{ 2 }+{ m }_{ 2 }{ v }_{ 2 }^{ 2 }.....................(2)$

The goal here is to find $v_2$ , according to (1)

${ v }_{ 1 }=\frac { v({ m }_{ 1 }-{ m }_{ 2 })-{ m }_{ 2 }v_{ 2 } }{ { m }_{ 1 } }.....................(3)$

Substitute (3) in (2)

${ m }_{ 1 }{ v }^{ 2 }+{ m }_{ 2 }{ v }^{ 2 }={ m }_{ 1 }{ (\frac { v({ m }_{ 1 }-{ m }_{ 2 })-{ m }_{ 2 }v_{ 2 } }{ { m }_{ 1 } } ) }^{ 2 }+{ m }_{ 2 }{ v }_{ 2 }$

Solving this equality for $v_2$ results in this quadratic equation

$({ m }_{ 1 }+{ m }_{ 2 }){ v }_{ 2 }^{ 2 }-2v({ m }_{ 1 }+{ m }_{ 2 }){ { v }_{ 2 } }+({ m }_{ 2 }-3{ m }_{ 1 }){ v }^{ 2 }=0$

Solving that equation

${ v }_{ 2 }=\frac { 2v({ m }_{ 1 }+{ m }_{ 2 })\pm \sqrt { 4{ v }^{ 2 }{ ({ m }_{ 1 }+{ m }_{ 2 }) }^{ 2 }-4{ v }^{ 2 }({ m }_{ 1 }+{ m }_{ 2 })({ m }_{ 2 }-3{ m }_{ 1 }) } }{ 2({ m }_{ 1 }+{ m }_{ 2 }) } =\frac { v({ m }_{ 1 }+{ m }_{ 2 })\pm v\sqrt { { ({ m }_{ 1 }+{ m }_{ 2 }) }(4{ m }_{ 1 }) } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { v({ m }_{ 1 }+{ m }_{ 2 })\pm 2v\sqrt { { ({ m }_{ 1 }+{ m }_{ 2 }) }{ m }_{ 1 } } }{ { m }_{ 1 }+{ m }_{ 2 } }$

${ v }_{ 2 }=v\pm 2v\sqrt { \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } } =v(1\pm 2\sqrt { \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } } )$

As $m_1>m_2$

${ v }_{ 2 }=v(1+2\sqrt { \frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } } } )$

If $m_1>>m_2$ , then $\frac { { m }_{ 1 } }{ { m }_{ 1 }+{ m }_{ 2 } }$ approaches 1 and $v_2$ approaches $3v$ . With this, after the bounce, the tennis ball will follow this equation about its height

$h(t)=-\frac { 1 }{ 2 } g{ t }^{ 2 }+3vt+d$

its maximum height is reached when $t=\frac { 3v }{ g }$ (see this for more information), with this

${ h }_{ max }=-\frac { 1 }{ 2 } g*\frac { 9{ v }^{ 2 } }{ { g }^{ 2 } } +3v*\frac { 3v }{ g } +d=\frac { 1 }{ 2 } (\frac { 9{ v }^{ 2 } }{ g } )+d=\quad \frac { 1 }{ 2 } (\frac { 9*2gh }{ g } )+d=\boxed { 9h+d }$