A Ball off a Roof Rises

Q. A ball is thrown straight up at 20 m/s from the top of a building 25 m tall. Taking g as 10 m/s², how high above the ground does it rise, in metres? expects 45.00 m

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The web of concepts

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Q. A ball is thrown straight up at 20 m/s $v_{0} = 20$ m/s from the top of a building 25 m $x_{0} = 25$ m tall. Taking g as $a = - g$ 10 m/s² $g = 10$ m/s², how high above the ground does it rise $v_{1} = 0$ m/s, in metres?

The playground

Inputs

throw speed

v_{0}

20.0 m/s

building height

x_{0}

25.00 m

gravity

g

10 m/s²

Outputs

height

x_{1}

The solution

Solved with UniformAcceleration

Given

$v_{0} = 20$ m/s
$x_{0} = 25$ m
$g = 10$ m/s²
$v_{1} = 0$ m/s from “how high above the ground does it rise”
$a = - g$ from “thrown straight up” and “Taking g as”

Find $x_{1}$ .

Working

thrown straight up and Taking g as $a = - g$
$a = - 10$
$a = −10$ m/s²
velocity after a time(rising) $v_{1} = v_{0} + a \cdot Δ t$
$0 = 20 + (−10) \cdot Δ t$
$Δ t = 2$ s
position after a time(rising) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$x_{1} = 25 + 20 \cdot 2 + \frac{(−10) \cdot 2^{2}}{2}$
$x_{1} = 45.00 m$

Conclusion

At the top the ball is still for an instant, so the speed it had is all spent against gravity over the height it climbs.

The Ruby

1require_relative "../physics"23question :a_ball_off_a_roof_rises do4  kind :numerical5  describes "A ball is thrown straight up at %{throw_speed} from the top of a building %{building_height} tall. Taking g as " \6          "%{gravity}, %{height}, in metres?"7  given throw_speed: "20 m/s"8  given building_height: "25 m"9  given gravity: "10 m/s2"10  asks height: "how high above the ground does it rise"11end

1require_relative "../questions/a_ball_off_a_roof_rises"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :a_ball_off_a_roof_rises, 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, building_height: ball.initial_position, gravity: g10  deduce_from "how high above the ground does it rise", ball.final_velocity => "0 m/s"11  take height: ball.final_position12  answer find_value(:height, in: "m")1314  conclusion "At the top the ball is still for an instant, so the speed it had is all spent " \15             "against gravity over the height it climbs."16end

1require_relative "../solutions/a_ball_off_a_roof_rises"23play_with :a_ball_off_a_roof_rises do4  title "A Ball off a Roof Rises"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 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