At the Top of a Throw

Q. A ball thrown straight up is at its highest point. Which are true? expects A, C

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

The playground

The solution

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 Ruby

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

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

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

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