At the Top of a Throw

solved UniformAcceleration

Q. A ball thrown straight up $a = - g$ is at its highest point $v_{1} = 0$ m/s. Which are true?

How it's solved

Solved with UniformAcceleration

Given

$v_{1} = 0$ m/s from “at its highest point”
$a = - g$ from “thrown straight up”

Assumed

$v_{0} = 20$ m/s
$g = 9.8$ m/s²

Find $a$ .

Working

thrown straight up $a = - g$
$a = - 9.8$
$a = −9.8$ m/s²
$v_{1} = 0$ : holds
$a = 0$ : does not hold
$a = - g$ : holds
$a = 0$ : does not hold

Conclusion

Gravity does not stop pulling at the top: the velocity passes through zero on its way from up to down, and the acceleration is g downward all the time, which is why the ball does not stay there.

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 :at_the_top_of_a_throw do4  kind :mcq5  describes "A ball thrown straight up is at its highest point. Which are true?"67  option "Its velocity is zero", as: :still8  option "Its acceleration is zero", as: :no_acceleration9  option "Its acceleration is g, downward", as: :g_down10  option "It is in balance for an instant", as: :balance11end

solution.rb

1require_relative "../questions/at_the_top_of_a_throw"2require_relative "../science/motion_in_a_straight_line/uniform_acceleration"34solve :at_the_top_of_a_throw, using: UniformAcceleration do5  objects :ball6  ball.goes(:rising)78  deduce_from "thrown straight up", ball.acceleration => -g9  deduce_from "at its highest point", ball.final_velocity => "0 m/s"10  assume ball.initial_velocity => "20 m/s"11  assume g => "9.8 m/s2"12  find_value ball.acceleration13  take still: ball.final_velocity == 0, no_acceleration: ball.acceleration == 0,14       g_down: ball.acceleration == -g, balance: ball.acceleration == 015  answer options.select { |o| holds?(o.taken) }1617  conclusion "Gravity does not stop pulling at the top: the velocity passes through zero on its " \18             "way from up to down, and the acceleration is g downward all the time, which is why " \19             "the ball does not stay there."20end

playground.rb

1require_relative "../solutions/at_the_top_of_a_throw"23play_with :at_the_top_of_a_throw do4  title "At the Top of a Throw"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