Projectile motion: two motions at once

Projectile motion

Why does a thrown ball follow a parabola, and which angle throws farthest?

Launch a ball at any speed, angle and height and watch its velocity split into a steady horizontal part and a vertical part that gravity changes. Read off flight time, range and peak height.

Drag the ringed points on the board.

High school
Independent components
Horizontal launch
Range and angle
Parabolic path

v_0

16.0

\theta\,(^\circ)

45°

h

0.0

t

1.20

R = \dfrac{v_0^2\sin 2\theta}{g}

$t$: 1.20
$t_{\text{flight}}$: 2.31
$\text{range}$: 26.12
$H_{\max}$: 6.53
$v_x$: 11.31
$v_y$: -0.45

Ballin the air

Guided tour

1Thrown sideways from a cliff
Throw horizontally at $10\,\text{m/s}$ from $h=20\,\text{m}$ . Falling takes the same time as a dropped ball, $t=\sqrt{2h/g}\approx2.02\,\text{s}$ , and in that time the ball travels $10\times2.02\approx20.2\,\text{m}$ sideways.
2Launched at 45°
From the ground at $12\,\text{m/s}$ and $45^\circ$ , both components are $12\cos45^\circ\approx8.49\,\text{m/s}$ . The ball is up for $1.73\,\text{s}$ , peaks at $3.67\,\text{m}$ and lands $14.69\,\text{m}$ away.
330°: lower and shorter
Drop the angle to $30^\circ$ . The path is flatter and the range shrinks to $\dfrac{v_0^2\sin60^\circ}{g}\approx12.73\,\text{m}$ .
460°: same landing spot
Now try $60^\circ$ . The ball goes much higher but lands at the same $12.73\,\text{m}$ : angles that add up to $90^\circ$ give the same range, and $45^\circ$ sits in the middle with the longest one.

Explanation

1
Two motions that ignore each other
Ignoring air resistance, the only force is gravity, and it points straight down. So the horizontal velocity never changes, while the vertical velocity changes by $g=9.8\,\text{m/s}^2$ every second:
$x=v_0\cos\theta\cdot t,\qquad y=h+v_0\sin\theta\cdot t-\tfrac12gt^2$
Watch the blue arrow on the board: it keeps its length the whole flight. Only the red one changes.
2
Why the path is a parabola
Eliminate $t$ using $t=\dfrac{x}{v_0\cos\theta}$ :
$y=h+x\tan\theta-\frac{g}{2v_0^2\cos^2\theta}\,x^2$
That is a quadratic in $x$ with a negative leading coefficient, a parabola opening downward. The constant horizontal speed turns "time" into "distance", and the $t^2$ of free fall becomes $x^2$ .
3
Range from level ground
With $h=0$ the ball lands when $y=0$ again, after $t=\dfrac{2v_0\sin\theta}{g}$ . Multiply by the horizontal speed:
$R=\frac{v_0^2\sin2\theta}{g}$
$\sin2\theta$ is largest at $2\theta=90^\circ$ , so $45^\circ$ throws farthest, and $\theta$ and $90^\circ-\theta$ give the same $\sin2\theta$ . From a height the best angle is a little below $45^\circ$ ; set $h$ and try it.

Check yourself

Q01

Two balls leave a table at the same moment: one is dropped, one is pushed off horizontally. Which lands first?

Q02

At the top of its path, what is the velocity of a ball launched at $v_0$ and angle $\theta$ ?

Q03

Launched from the ground at $20\,\text{m/s}$ and $30^\circ$ ( $g=10$ ), what is the range?

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