How High It Goes

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, how high does it rise $v_{1} = 0$ m/s, in metres?

Inputs

throw speed

v_{0}

29.4 m/s

gravity

g

9.8 m/s²

Outputs

height

x_{1}

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 29.4$ m/s
$g = 9.8$ m/s²
$v_{1} = 0$ m/s from “how high does it rise”
$a = - g$ from “thrown straight up” and “Taking g as”

Assumed

$x_{0} = 0$ m

Find $x_{1}$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 9.8$
$a = −9.8$ m/s²
velocity after a time(rising) $v_{1} = v_{0} + a \cdot Δ t$
$0 = 29.4 + (−9.8) \cdot Δ t$
$Δ t = 3$ s
position after a time(rising) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = 0 + 29.4 \cdot 3 + \frac{(−9.8) \cdot 3^{2}}{2}$
$x_{1} = 44.10 m$

Conclusion

The height goes as the square of the launch speed: throw twice as fast and it climbs four times as high.

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_high_it_goes do4  kind :numerical5  describes "A ball is thrown straight up at %{throw_speed}. Taking g as %{gravity}, %{height}, in metres?"6  given throw_speed: "29.4 m/s"7  given gravity: "9.8 m/s2"8  asks height: "how high does it rise"9end

solution.rb

1require_relative "../questions/how_high_it_goes"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :how_high_it_goes, using: UniformAcceleration do5  objects :ball6  ball.goes(:rising)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 "how high does it rise", ball.final_velocity => "0 m/s"12  take height: ball.final_position13  answer find_value(:height, in: "m")1415  conclusion "The height goes as the square of the launch speed: throw twice as fast and it " \16             "climbs four times as high."17end

playground.rb

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

v_{av} = \frac{Δ x}{Δ t}

displacement a time(rising)

c_{av} = \frac{d}{Δ t}

path a time(rising)

Δ v = v_{1} - v_{0}

velocity gained(rising)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(rising)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(rising)

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

position after a time(rising)

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

velocity after a distance(rising)

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

mean of the velocities(rising)

a = - g

thrown straight up and Taking g as