How Soon a Stone Reaches Fifteen Metres

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 does it reach 15 m $x_{1} = 15$ m above the point of release $x_{0} = 0$ m?

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”

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 earliest time is taken
$Δ t = 1.00 s$

Conclusion

The stone is at that height twice, going up and coming down; it reaches it the first time, on the way up.

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 :how_soon_a_stone_reaches_fifteen_metres 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?"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 does it reach"11end

solution.rb

1require_relative "../questions/how_soon_a_stone_reaches_fifteen_metres"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :how_soon_a_stone_reaches_fifteen_metres, 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  take time: flying.time_taken12  answer find_value(:time, in: "s")1314  conclusion "The stone is at that height twice, going up and coming down; it reaches it the first time, on the " \15             "way up."16end

playground.rb

1require_relative "../solutions/how_soon_a_stone_reaches_fifteen_metres"23play_with :how_soon_a_stone_reaches_fifteen_metres do4  title "How Soon a Stone Reaches Fifteen Metres"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