Fifteen Metres Up on the Way Down

solved UniformAcceleration

Q. A stone is thrown straight up at 20 m/s $v_{0} = 20$ m/s. Taking g as $a = - g$ 10 m/s² $g = 10$ m/s², how many seconds after the throw is it 15 m $x_{1} = 15$ m above the point of release $x_{0} = 0$ m on its way down?

Inputs

throw speed

v_{0}

20.0 m/s

gravity

g

10 m/s²

height

x_{1}

15.00 m

Outputs

time

Δ t

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 20$ m/s
$g = 10$ m/s²
$x_{1} = 15$ m
$x_{0} = 0$ m from “above the point of release”
$a = - g$ from “thrown straight up” and “Taking g as”
$Δ t$ : the largest of the values that fit from “on its way down”

Find $Δ t$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 10$
$a = −10$ m/s²
position after a time(flying) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$15 = 0 + 20 \cdot Δ t + \frac{(−10) \cdot {Δ t}^{2}}{2}$
$Δ t = 1, 3$ sthe largest is taken
$Δ t = 3.00 s$

Conclusion

The stone passes the same height twice, once going up and once coming down; the later of the two times is the one on the way down.

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 :fifteen_metres_up_on_the_way_down do4  kind :numerical5  describes "A stone is thrown straight up at %{throw_speed}. Taking g as %{gravity}, %{time} %{height} above the " \6            "point of release on its way down?"7  given throw_speed: "20 m/s"8  given gravity: "10 m/s2"9  given height: "15 m"10  asks time: "how many seconds after the throw is it"11end

solution.rb

1require_relative "../questions/fifteen_metres_up_on_the_way_down"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :fifteen_metres_up_on_the_way_down, using: UniformAcceleration do5  objects :stone6  stone.goes(:flying)78  deduce_from "thrown straight up", "Taking g as", stone.acceleration => -g9  take throw_speed: stone.initial_velocity, gravity: g, height: stone.final_position10  deduce_from "above the point of release", stone.initial_position => "0 m"11  deduce_from "on its way down", flying.time_taken => :largest12  take time: flying.time_taken13  answer find_value(:time, in: "s")1415  conclusion "The stone passes the same height twice, once going up and once coming down; the later of the " \16             "two times is the one on the way down."17end

playground.rb

1require_relative "../solutions/fifteen_metres_up_on_the_way_down"23play_with :fifteen_metres_up_on_the_way_down do4  title "Fifteen Metres Up on the Way Down"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 stone during flying	$Δ x$	m	—
position of stone at end of flying	$x_{1}$	m	—
position of stone at start of flying	$x_{0}$	m	—
average velocity of stone during flying	$v_{av}$	m/s	—
time taken of flying	$Δ t$	s	0 … 1e+05
average speed of stone during flying	$c_{av}$	m/s	—
path length of stone during flying	$d$	m	—
change of velocity of stone during flying	$Δ v$	m/s	—
velocity of stone at end of flying from ground	$v_{1}$	m/s	—
velocity of stone at start of flying from ground	$v_{0}$	m/s	—
average acceleration of stone during flying	$a_{av}$	m/s²	—
acceleration of stone 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