A Ball off a Roof Rises

solved UniformAcceleration

Q. A ball is thrown straight up at 20 m/s $v_{0} = 20$ m/s from the top of a building 25 m $x_{0} = 25$ m tall. Taking g as $a = - g$ 10 m/s² $g = 10$ m/s², how high above the ground does it rise $v_{1} = 0$ m/s, in metres?

Inputs

throw speed

v_{0}

20.0 m/s

building height

x_{0}

25.00 m

gravity

g

10 m/s²

Outputs

height

x_{1}

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 20$ m/s
$x_{0} = 25$ m
$g = 10$ m/s²
$v_{1} = 0$ m/s from “how high above the ground does it rise”
$a = - g$ from “thrown straight up” and “Taking g as”

Find $x_{1}$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 10$
$a = −10$ m/s²
velocity after a time(rising) $v_{1} = v_{0} + a \cdot Δ t$
$0 = 20 + (−10) \cdot Δ t$
$Δ t = 2$ s
position after a time(rising) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = 25 + 20 \cdot 2 + \frac{(−10) \cdot 2^{2}}{2}$
$x_{1} = 45.00 m$

Conclusion

At the top the ball is still for an instant, so the speed it had is all spent against gravity over the height it climbs.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_ball_off_a_roof_rises do4  kind :numerical5  describes "A ball is thrown straight up at %{throw_speed} from the top of a building %{building_height} tall. Taking g as " \6          "%{gravity}, %{height}, in metres?"7  given throw_speed: "20 m/s"8  given building_height: "25 m"9  given gravity: "10 m/s2"10  asks height: "how high above the ground does it rise"11end

solution.rb

1require_relative "../questions/a_ball_off_a_roof_rises"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :a_ball_off_a_roof_rises, using: UniformAcceleration do5  objects :ball6  ball.goes(:rising)78  deduce_from "thrown straight up", "Taking g as", ball.acceleration => -g9  take throw_speed: ball.initial_velocity, building_height: ball.initial_position, gravity: g10  deduce_from "how high above the ground does it rise", ball.final_velocity => "0 m/s"11  take height: ball.final_position12  answer find_value(:height, in: "m")1314  conclusion "At the top the ball is still for an instant, so the speed it had is all spent " \15             "against gravity over the height it climbs."16end

playground.rb

1require_relative "../solutions/a_ball_off_a_roof_rises"23play_with :a_ball_off_a_roof_rises do4  title "A Ball off a Roof Rises"5end

The laws it drew on

quantity	written	units	within	from
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
velocity	$v$	m/s	—	Acceleration
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
acceleration	$a$	m/s²	—	UniformAcceleration
gravity	$g$	m/s²	—	UniformAcceleration
time	$t$	s	0 … 1e+05	UniformAcceleration
displacement of ball during rising	$Δ x$	m	—
position of ball at end of rising	$x_{1}$	m	—
position of ball at start of rising	$x_{0}$	m	—
average velocity of ball during rising	$v_{av}$	m/s	—
time taken of rising	$Δ t$	s	0 … 1e+05
average speed of ball during rising	$c_{av}$	m/s	—
path length of ball during rising	$d$	m	—
change of velocity of ball during rising	$Δ v$	m/s	—
velocity of ball at end of rising from ground	$v_{1}$	m/s	—
velocity of ball at start of rising from ground	$v_{0}$	m/s	—
average acceleration of ball during rising	$a_{av}$	m/s²	—
acceleration of ball during rising	$a$	m/s²	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

Δ x = x_{1} - x_{0}

change of position(rising)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(rising)

c_{av} = \frac{d}{Δ t}

path a time(rising)

Δ v = v_{1} - v_{0}

velocity gained(rising)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(rising)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(rising)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(rising)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(rising)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(rising)

a = - g

thrown straight up and Taking g as