Both Times a Ball Passes a Ledge

solved UniformAcceleration

Q. A ball is thrown straight up at 20 m/s $v_{0} = 20$ m/s from the foot of a wall $x_{0} = 0$ m with a ledge 15 m $x_{1} = 15$ m up it. Taking g as $a = - g$ 9.8 m/s² $g = 9.8$ m/s², at what times after the throw does it pass the ledge, in seconds?

Inputs

throw speed

v_{0}

20.0 m/s

height

x_{1}

15.00 m

gravity

g

9.8 m/s²

Outputs

times

Δ t

How it's solved

Solved with UniformAcceleration

Given

$v_{0} = 20$ m/s
$g = 9.8$ m/s²
$x_{1} = 15$ m
$x_{0} = 0$ m from “from the foot of a wall”
$a = - g$ from “thrown straight up” and “Taking g as”

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) $x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}$
$15 = 0 + 20 \cdot Δ t + \frac{(−9.8) \cdot {Δ t}^{2}}{2}$
$Δ t = 0.990241822, 3.09139083$ sall of them are taken
$Δ t = 0.99 s and 3.09 s$
$Δ t \approx 0.990, 3.09$ sto 3 significant figures

Conclusion

The ball passes the ledge twice, once rising and once falling back, so both times answer the question; they sit evenly either side of the top of the flight, about 2.04 s after the throw.

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 :both_times_a_ball_passes_a_ledge do4  kind :numerical5  describes "A ball is thrown straight up at %{throw_speed} from the foot of a wall with a ledge %{height} up it. " \6            "Taking g as %{gravity}, %{times}, in seconds?"7  given throw_speed: "20 m/s"8  given height: "15 m"9  given gravity: "9.8 m/s2"10  asks times: "at what times after the throw does it pass the ledge"11end

solution.rb

1require_relative "../questions/both_times_a_ball_passes_a_ledge"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :both_times_a_ball_passes_a_ledge, 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: g, height: ball.final_position10  deduce_from "from the foot of a wall", ball.initial_position => "0 m"11  take times: flying.time_taken12  answer find_value(:times, in: "s", figures: 3, root: :all)1314  conclusion "The ball passes the ledge twice, once rising and once falling back, so both times answer the " \15             "question; they sit evenly either side of the top of the flight, about 2.04 s after the throw."16end

playground.rb

1require_relative "../solutions/both_times_a_ball_passes_a_ledge"23play_with :both_times_a_ball_passes_a_ledge do4  title "Both Times a Ball Passes a Ledge"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