Fifteen Metres Up on the Way Down

Q. A stone is thrown straight up at 20 m/s. Taking g as 10 m/s², how many seconds after the throw is it 15 m above the point of release on its way down? expects 3.00 s

What the draft establishes

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The web of concepts

given read from the words equation asked

The equations it used

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Check an answer

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?

The playground

Inputs

throw speed

v_{0}

20.0 m/s

gravity

g

10 m/s²

height

x_{1}

15.00 m

Outputs

time

Δ t

The solution

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 Ruby

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

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

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

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