## Question:

A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m/s. At the same instant another ball B is thrown upward from the ground with an initial velocity of 20 m/s. Determine the height from the ground and the time at which they pass.

## Step-by-step

Origin at roof:

Ball A:

\begin{aligned} (+ \uparrow) \quad\quad & s = { s }_{ 0 } + { v }_{ 0 }t + \frac { 1 } { 2 } { a }_{ c } { t }^{ 2 } \\ & -s = 0 + 5t – \frac { 1 } { 2 } (9.81){ t }^{ 2 } \end{aligned}

Ball B:

\begin{aligned} (+ \uparrow) \quad\quad & s = { s }_{ 0 } + { v }_{ 0 }t + \frac { 1 } { 2 } { a }_{ c } { t }^{ 2 }\\ & -s = -30 + 20t – \frac { 1 } { 2 } (9.81){ t }^{ 2 } \end{aligned}

Solving,

$t = 2 s \\ s = 9.62m$

Distance from ground,

$d = (30 – 9.62) = 20.4m$

Also, origin at ground,

$s = { s }_{ 0 } + { v }_{ 0 }t + \frac { 1 } { 2 } { a }_{ c } { t }^{ 2 } \\ { s }_{ A } = 30 + 5t + \frac { 1 } { 2 }(-9.81){ t }^{ 2 } \\ { s }_{ B } = 0 + 20t + \frac { 1 } { 2 }(-9.81){ t }^{ 2 }$

Require

${ s }_{ A } = { s }_{ B } \\ 30 + 5t + \frac { 1 } { 2 }(-9.81){ t }^{ 2 } = 20t + \frac { 1 } { 2 }(-9.81){ t }^{ 2 } \\ t = 2s \\ { s }_{ B } = 20.4m$