A Ball off a Roof Lands

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 many seconds pass before it hits the ground $x_{1} = 0$ m?

Inputs

throw speed

v_{0}

20.0 m/s

building height

x_{0}

25.00 m

gravity

g

10 m/s²

Outputs

time

Δ t

How it's solved

Solved with UniformAcceleration

Given

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

Find $Δ t$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 10$
$a = −10$ m/s²
position after a time(flying) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$0 = 25 + 20 \cdot Δ t + \frac{(−10) \cdot {Δ t}^{2}}{2}$
$Δ t = 5.00 s$

Conclusion

There is no need to split the flight at the top: the acceleration never changes, so one equation covers the way up and the way down. It has two roots, and only the one after the throw is an answer.

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_lands 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}, %{time}?"7  given throw_speed: "20 m/s"8  given building_height: "25 m"9  given gravity: "10 m/s2"10  asks time: "how many seconds pass before it hits the ground"11end

solution.rb

1require_relative "../questions/a_ball_off_a_roof_lands"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :a_ball_off_a_roof_lands, using: UniformAcceleration do5  objects :ball6  ball.goes(:flying)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 "hits the ground", ball.final_position => "0 m"11  take time: flying.time_taken12  answer find_value(:time, in: "s")1314  conclusion "There is no need to split the flight at the top: the acceleration never changes, so " \15             "one equation covers the way up and the way down. It has two roots, and only the one " \16             "after the throw is an answer."17end

playground.rb

1require_relative "../solutions/a_ball_off_a_roof_lands"23play_with :a_ball_off_a_roof_lands do4  title "A Ball off a Roof Lands"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 flying	$Δ x$	m	—
position of ball at end of flying	$x_{1}$	m	—
position of ball at start of flying	$x_{0}$	m	—
average velocity of ball during flying	$v_{av}$	m/s	—
time taken of flying	$Δ t$	s	0 … 1e+05
average speed of ball during flying	$c_{av}$	m/s	—
path length of ball during flying	$d$	m	—
change of velocity of ball during flying	$Δ v$	m/s	—
velocity of ball at end of flying from ground	$v_{1}$	m/s	—
velocity of ball at start of flying from ground	$v_{0}$	m/s	—
average acceleration of ball during flying	$a_{av}$	m/s²	—
acceleration of ball during flying	$a$	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(flying)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(flying)

c_{av} = \frac{d}{Δ t}

path a time(flying)

Δ v = v_{1} - v_{0}

velocity gained(flying)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(flying)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(flying)

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

position after a time(flying)

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

velocity after a distance(flying)

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

mean of the velocities(flying)

a = - g

thrown straight up and Taking g as