Horizontal and oblique projectile motion

What is it about?

A thrown ball performs two motions at the same time that do not affect each other (principle of superposition):

horizontal: a uniform motion with the constant speed $v_x = v_0 \cdot \cos\alpha$ ,
vertical: a vertical throw with $v_{y,0} = v_0 \cdot \sin\alpha$ , slowed down by the gravitational acceleration $g$ : $v_y(t) = v_0 \sin\alpha - g\,t$ .

x(t) = v_0 \cos\alpha \cdot t \qquad y(t) = h_0 + v_0 \sin\alpha \cdot t - \tfrac{1}{2} g t^2

Eliminating $t$ gives the trajectory – a parabola:

y = h_0 + \tan\alpha \cdot x - \frac{g}{2 v_0^2 \cos^2\alpha} \cdot x^2

For a launch from the ground ( $h_0 = 0$ ), time of flight, range and maximum height are:

T = \frac{2 v_0 \sin\alpha}{g} \qquad w = \frac{v_0^2 \sin(2\alpha)}{g} \qquad h_{\max} = \frac{v_0^2 \sin^2\alpha}{2g}

The range is then largest at $\alpha = 45^\circ$ . Two angles that add up to $90^\circ$ (e.g. $30^\circ$ and $60^\circ$ ) give the same range. If you throw from a height $h_0 > 0$ , the best angle is below $45^\circ$ . In a horizontal throw ( $\alpha = 0^\circ$ ) the flight lasts exactly as long as a free fall from $h_0$ : $T = \sqrt{2h_0/g}$ . The range is then $w = v_0 \cdot \sqrt{2h_0/g}$ .

The speed combines both components: $v = \sqrt{v_x^2 + v_y^2}$ . Without air resistance the ball lands with $v = \sqrt{v_0^2 + 2 g h_0}$ – this also follows from conservation of energy.

Try it

Drag the tip of the orange arrow to change $v_0$ and $\alpha$ , then press Throw. Drag the launch point upwards to change $h_0$ .
Open the components view. The blue shadow on the ground moves uniformly, the red one on the vertical rail moves like a vertical throw. Drag in a graph to move through time. Components
Which angle gives the largest range? Tap different angles in the graph. Compare angles
Throw at $v_0 = 10\,\tfrac{\text{m}}{\text{s}}$ from $h_0 = 20\,\text{m}$ . Where is the maximum now? (At about $24^\circ$ .) From 20 m
Hit the target! After the first hit the simulation shows both solution angles. Target
Compare a football, a table tennis ball and a shot put with and without air. Air resistance
On the Moon ( $g = 1.62\,\tfrac{\text{m}}{\text{s}^2}$ ) the same ball flies about 6 times as far: $19.86\,\text{m}$ become $120.3\,\text{m}$ . Moon

Tasks

Task 1: Calculating the range

Load task

A ball is thrown from the ground at $v_0 = 15\,\tfrac{\text{m}}{\text{s}}$ at $45^\circ$ ( $g = 9.81\,\tfrac{\text{m}}{\text{s}^2}$ ). Calculate the time of flight, the range and the maximum height.

Show solution

$T = \frac{2 \cdot 15 \cdot \sin 45^\circ}{9.81}\,\text{s} \approx 2.16\,\text{s} \qquad w = \frac{15^2 \cdot \sin 90^\circ}{9.81}\,\text{m} \approx 22.9\,\text{m} \qquad h_{\max} = \frac{15^2 \cdot \sin^2 45^\circ}{2 \cdot 9.81}\,\text{m} \approx 5.73\,\text{m}$

Task 2: Same range

Show that $30^\circ$ and $60^\circ$ give the same range at $v_0 = 15\,\tfrac{\text{m}}{\text{s}}$ . Which path is higher, which one takes longer? Show comparison

Show solution

$\sin(2 \cdot 30^\circ) = \sin 60^\circ = \sin 120^\circ = \sin(2 \cdot 60^\circ)$ , so $w \approx 19.86\,\text{m}$ for both. The $60^\circ$ throw is three times as high ( $8.60\,\text{m}$ instead of $2.87\,\text{m}$ , because $\sin^2 60^\circ : \sin^2 30^\circ = 3$ ) and takes $\sqrt{3}$ times as long ( $2.65\,\text{s}$ instead of $1.53\,\text{s}$ ).

Task 3: Horizontal throw from a tower

Load task

A stone is thrown horizontally at $10\,\tfrac{\text{m}}{\text{s}}$ from a $20\,\text{m}$ high tower. When and where does it land? How fast is it then?

Show solution

$T = \sqrt{2 \cdot 20 / 9.81}\,\text{s} \approx 2.02\,\text{s}$ , $w = 10 \cdot 2.02\,\text{m} \approx 20.2\,\text{m}$ . On impact $v_y = 9.81 \cdot 2.02\,\tfrac{\text{m}}{\text{s}} \approx 19.8\,\tfrac{\text{m}}{\text{s}}$ , so

$v = \sqrt{10^2 + 19.8^2}\,\tfrac{\text{m}}{\text{s}} \approx 22.2\,\tfrac{\text{m}}{\text{s}}.$

Task 4: Hitting the target

Load task

A target hangs $18\,\text{m}$ away at a height of $2\,\text{m}$ . The ball starts on the ground at $v_0 = 15\,\tfrac{\text{m}}{\text{s}}$ . At which angles does it hit? Put $x = 18\,\text{m}$ , $y = 2\,\text{m}$ into the trajectory equation and solve for $\tan\alpha$ (using $\tfrac{1}{\cos^2\alpha} = 1 + \tan^2\alpha$ ).

Show solution

$2 = 18\tan\alpha - \frac{9.81 \cdot 18^2}{2 \cdot 15^2}(1 + \tan^2\alpha)$ leads to $7.063\tan^2\alpha - 18\tan\alpha + 9.063 = 0$ . Solutions: $\tan\alpha \approx 0.691$ or $1.858$ , so $\alpha \approx 34.6^\circ$ (flat path) and $\alpha \approx 61.7^\circ$ (steep path). With less than about $14.05\,\tfrac{\text{m}}{\text{s}}$ the target would be out of reach. Show solution

Task 5: Air resistance

Compare the path of a football ( $v_0 = 28\,\tfrac{\text{m}}{\text{s}}$ , $\alpha = 35^\circ$ ) with and without air. Name three differences and explain the energy graph. Air resistance

Show solution

Without air: $w = \frac{28^2 \cdot \sin 70^\circ}{9.81}\,\text{m} \approx 75.1\,\text{m}$ . With air the ball only flies about $45.7\,\text{m}$ and not as high; the path is no longer a parabola, the apex lies further along, and the ball comes down more steeply. The best angle is about $41^\circ$ , i.e. below $45^\circ$ . The energy per kilogram $\tfrac12 v^2 + g\,y$ stays constant without air; with air it decreases – it is converted into internal energy of the air and the ball.

Task 6: Horizontal throw on the Moon

Load task

The same stone as in Task 3 is thrown horizontally on the Moon ( $g = 1.62\,\tfrac{\text{m}}{\text{s}^2}$ ) at $10\,\tfrac{\text{m}}{\text{s}}$ from a height of $20\,\text{m}$ . Calculate the time of flight, the range and the impact speed. By what factor is the range larger than on Earth?

Show solution

$T = \sqrt{2 \cdot 20 / 1.62}\,\text{s} \approx 4.97\,\text{s}$ , $w = 10 \cdot 4.97\,\text{m} \approx 49.7\,\text{m}$ . On impact $v_y = 1.62 \cdot 4.97\,\tfrac{\text{m}}{\text{s}} \approx 8.05\,\tfrac{\text{m}}{\text{s}}$ , so $v = \sqrt{10^2 + 8.05^2}\,\tfrac{\text{m}}{\text{s}} \approx 12.8\,\tfrac{\text{m}}{\text{s}}$ . The range is larger by the factor $\sqrt{9.81/1.62} \approx 2.46$ than on Earth ( $20.2\,\text{m}$ ).

Ideas for the classroom

Use: mechanics, superposition of motions, grade 10 or 11 depending on the curriculum. The components view works well as an introduction, the target mode as practice or as a group challenge.
Representation: stroboscope dots and shadows show that the horizontal spacing stays the same while the vertical spacing does not. With slow motion and dragging in the graphs you can discuss individual moments.
Model: point mass, constant $g$ ; air resistance quadratic in $v$ with approximate values (football $m = 0.43\,\text{kg}$ , $d = 22\,\text{cm}$ , $c_w \approx 0.25$ ; table tennis ball $2.7\,\text{g}$ ; shot put $7.26\,\text{kg}$ ). There is no air on the Moon, and on Mars it is about 60 times thinner.
Common stumbling blocks:
- “At the highest point the speed is zero.” – Only $v_y$ is zero, $v_x$ stays the same.
- “A fast ball thrown horizontally falls more slowly.” – The time of fall depends only on $h_0$ .
- “ $45^\circ$ is always best.” – Only for $h_0 = 0$ and without air resistance.