Free fall

Vertical motion under gravity: dropping, throwing up and down, graphs of acceleration, velocity and coordinate.

Terms

Studied 0 of 3

Formulas

Velocity

v_y = v_{0y} - g\,t

Quantities
Symbol	Name	Unit
$v_y$	velocity component on axis $Oy$ (up)	m/s
$v_{0y}$	initial velocity	m/s
$g$	acceleration due to gravity	m/s²
$t$	time	s

Note. The $Oy$ axis points up, hence the minus before $g$ . A drop from rest: speed $v = g\,t$ .

Coordinate (height)

y = y_0 + v_{0y}\,t - \dfrac{g t^2}{2}

Quantities
Symbol	Name	Unit
$y$	height at time $t$	m
$y_0$	initial height	m
$v_{0y}$	initial velocity	m/s
$g$	acceleration due to gravity	m/s²
$t$	time	s

Note. Landing means $y = 0$ . Use this condition to find the time of fall.

Drop from rest

h = \dfrac{g t^2}{2}

Quantities
Symbol	Name	Unit
$h$	height fallen	m
$g$	acceleration due to gravity	m/s²
$t$	time of fall	s

Solved for another quantity:

t = \sqrt{\dfrac{2h}{g}}

v = g\,t

v = \sqrt{2 g h}

Throwing up: the rise

t_{\text{подъёма}} = \dfrac{v_0}{g}, \qquad H_{\max} = \dfrac{v_0^2}{2g}

Quantities
Symbol	Name	Unit
$t_{\text{подъёма}}$	time to reach the top	s
$H_{\max}$	maximum height above the throw point	m
$v_0$	initial speed (upward)	m/s
$g$	acceleration due to gravity	m/s²

Note. Both follow from $v_y = 0$ at the top. Falling back from that height takes as long as the rise.

Lab

Set the quantities with the sliders. The graphs and the formulas with your numbers update at once.

Motion of the body

Drag the ball with the mouse to set its starting height. The dots mark the body's position every 0.5 s.

t

= 0 sPlayback speed

Graphs

The graphs are drawn as the body moves. Click a graph to move through the recording.

y(t)

s_y(t)

$s_y = y - y_0$ . It decreases when the body moves against the axis.

l(t)

Distance never decreases: after a turnaround it grows while $s_y$ shrinks.

v_y(t)

The shaded area is the displacement $s_y$ .

a_y(t)

$a_y = -g$ : below the axis because $Oy$ points up. Zero on the ground.

t

0 s

v_y

0 m/s

s_y

0 m

l

0 m

Formulas with your numbers

v_y = v_{0y} - g t = 0 - 10\cdot 0 = 0\ \text{m/s}

y = y_0 + v_{0y}t - \dfrac{g t^2}{2} = 45 + 0\cdot 0 - \dfrac{10\cdot 0^{2}}{2} = 45\ \text{m}

Lab «Cannon»

Hit the target: throw a body up or horizontally, watch the dots every 0.5 s and the velocity components — the horizontal and vertical motions are independent.

Open full screen →

How to read the graphs

The $Oy$ axis points up, so $a_y = -g$ is negative: the $a_y(t)$ graph lies below the time axis.
$v_y(t)$ is a line with slope $-g$ . Going up the velocity is positive and falling, at the top it is zero, going down it is negative.
$y(t)$ is a downward-opening parabola. Its vertex is the maximum height (if the body was thrown up).
The graphs end when the body reaches the ground, at $y = 0$ .

Loading…