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Ball A is thrown vertically upwards with a velocity of { v }_{ 0 }. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t < 2{ v }_{ 0 }/g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

Step-by-step

Kinematics: First, we will consider the motion of ball A with { ({ v  }_{ A }) }_{ 0 } = { v }_{ 0 }, { ({ s }_{ A }) }_{ 0 } = 0, { s }_{ A } = h, { t }_{ A } = { t }^{ ‘ }, \text{ and } { ({ a }_{ c }) }_{ A } = -g.

\begin{aligned} (+ \uparrow) \quad\quad & { s }_{ A } = { ({ s }_{ A }) }_{0} + { ({ v }_{ A }) }_{ 0 } { t }_{ A } + \frac { 1 } { 2 } { ({ a }_{ c }) }_{ A } { { t }_{ A } }^{ 2 } \\ & h = 0 + { v }_{ 0 } { t }^{ ‘ } + \frac { 1 } { 2 } (-g) { ({ t }^{ ‘ }) }^{ 2 } \\ & h = { v }_{ 0 } { t }^{ ‘ } – \frac { g } { 2 } { { t }^{ ‘ } }^{ 2 } \quad\quad\quad\quad\quad (1) \end{aligned} \\ \\ \begin{aligned} (+ \uparrow) \quad\quad & { v }_{ A } = { ({ v }_{ A }) }_{ 0 } + { ({ a }_{ c }) }_{ A } { t }_{ A } \\ & { v }_{ A } = { v }_{ 0 } + (-g)({ t }^{ ‘ }) \\ & { v }_{ A } = { v }_{ 0 } – g{ t }^{ ‘ } \quad\quad\quad (2) \end{aligned}

The motion of ball B requires { ({ v }_{ B }) }_{ 0 } = { v }_{ 0 }, { ({ s }_{ B }) }_{ 0 } = 0, { s }_{ B } = h, { t }_{ B } = \tilde{ t } – t, \text{ and } { ({ a }_{ c }) }_{ B } = -g.

\begin{aligned} (+ \uparrow) \quad\quad & { s }_{ B } = { ({ s }_{ B }) }_{ 0 } + { ({ v }_{ B }) }_{ 0 } { t }_{ B } + \frac { 1 } { 2 } ({ a }_{ c })_{ B } { { t }_{ B } }^{ 2 } \\ & h = 0 + { v }_{ 0 } ({ t }^{ ‘ } – t) + \frac { 1 } { 2 } (-g){ ({ t }^{ ‘ } – 2) }^{ 2 } \\ & h = { v }_{ 0 }({ t }^{ ‘ } – t) – \frac { g } { 2 } { ({ t }^{ ‘ } – t) }^{ 2 } \quad\quad\quad (3) \end{aligned} \\ \\ \begin{aligned} (+ \uparrow) \quad\quad & { v }_{ B } = { ({ v }_{ B }) }_{ 0 } + { ({ a }_{ c }) }_{ B } { t }_{ B } \\ & { v }_{ B } = { v }_{ 0 } + (-g)({ t }^{ ‘ } – t) \\ & { v }_{ B } = { v }_{ 0 } – g({ t }^{ ‘ } – t) \quad\quad\quad (4) \end{aligned}

Solving Eqs. (1) and (3),

{ v }_{ 0 } { t }^{ ‘ } – \frac { g } { 2 } { { t }^{ ‘ } }^{ 2 } = { v }_{ 0 } ({ t }^{ ‘ } – t) – \frac { g } { 2 } { ({ t }^{ ‘ } – t) }^{ 2 } \\ { t }^{ ‘ } = \frac { 2{ v }_{ 0 } + gt } { 2g }

Substituting this result into Eqs. (2) and (4),

\begin{aligned} { v }_{ A } &= { v }_{ 0 } – g (\frac { 2{ v }_{ 0 } + gt } { 2g }) \\ &= -\frac { 1 } { 2 }gt = \frac { 1 } { 2 }gt \downarrow \end{aligned} \\ \\ \begin{aligned}{ v }_{ B } &= { v }_{ 0 } – g(\frac { 2{ v }_{ 0 } + gt } { 2g } – t) \\ &= \frac { 1 } { 2 }gt \uparrow \end{aligned}

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