Back in the Hands

solved UniformAcceleration

Q. A ball is thrown straight up at 29.4 m/s $v_{0} = 29.4$ m/s. Taking g as $a = - g$ 9.8 m/s² $g = 9.8$ m/s², after how many seconds is it back in the thrower's hands $x_{1} = 0$ m?

Inputs

throw speed

v_{0}

29.4 m/s

gravity

g

9.8 m/s²

Outputs

time

Δ t

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 29.4$ m/s
$g = 9.8$ m/s²
$x_{1} = 0$ m from “back in the thrower's hands”
$a = - g$ from “thrown straight up” and “Taking g as”

Assumed

$x_{0} = 0$ m

Find $Δ t$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 9.8$
$a = −9.8$ m/s²
position after a time(flying) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$0 = 0 + 29.4 \cdot Δ t + \frac{(−9.8) \cdot {Δ t}^{2}}{2}$
$Δ t = 0, 6$ sthe start is set aside, as a later time fits
$Δ t = 6.00 s$

Conclusion

It takes as long to come down as to go up, and comes back at the speed it left with, pointing the other way.

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 :back_in_the_hands do4  kind :numerical5  describes "A ball is thrown straight up at %{throw_speed}. Taking g as %{gravity}, %{time}?"6  given throw_speed: "29.4 m/s"7  given gravity: "9.8 m/s2"8  asks time: "after how many seconds is it back in the thrower's hands"9end

solution.rb

1require_relative "../questions/back_in_the_hands"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :back_in_the_hands, 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, gravity: g10  assume ball.initial_position => "0 m"11  deduce_from "back in the thrower's hands", ball.final_position => "0 m"12  take time: flying.time_taken13  answer find_value(:time, in: "s")1415  conclusion "It takes as long to come down as to go up, and comes back at the speed it left " \16             "with, pointing the other way."17end

playground.rb

1require_relative "../solutions/back_in_the_hands"23play_with :back_in_the_hands do4  title "Back in the Hands"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