Projectile motion.

With the fan on, a constant force acts on the cart, so its acceleration is constant too: $a = \dfrac{F}{m} = \dfrac{0.5}{1} = 0.5\ \mathrm{m/s^2}$$a =$$\dfrac{F}{m} =$$\dfrac{0.5}{1} =$$0.5\ \mathrm{m/s^2}$⁠. From rest, the distance after time $t$⁠ is $x = \tfrac12 a t^2 = 0.25\, t^2$$x =$$\tfrac12 a t^2 =$$0.25\, t^2$⁠, so after $0.5$⁠, $1$⁠, $1.5$⁠ and $2$⁠ seconds the cart is at $0.0625$⁠, $0.25$⁠, $0.5625$⁠ and $1$⁠ metre.

The gaps between the marks are the differences: $0.0625$⁠, $0.1875$⁠, $0.3125$⁠ and $0.4375$⁠ metres, in the ratio $1 : 3 : 5 : 7$⁠. Each gap is bigger than the one before by the same amount, which is the signature of constant acceleration.

First the acceleration: $a = \dfrac{F}{m}$⁠. Then the distance from rest: $x = \tfrac12 a t^2$⁠.

1. The acceleration, from Newton’s second law: $a = \dfrac{F}{m} = \dfrac{0.5\ \mathrm{N}}{1\ \mathrm{kg}} = 0.5\ \mathrm{m/s^2}$$a =$$\dfrac{F}{m} =$$\dfrac{0.5\ \mathrm{N}}{1\ \mathrm{kg}} =$$0.5\ \mathrm{m/s^2}$⁠.
2. The cart starts from rest, so $x = \tfrac12 a t^2$⁠.
3. Put in $t = 2\ \mathrm{s}$⁠: $x = \tfrac12 \cdot 0.5 \cdot 2^2 = 1\ \mathrm{m}$$x =$$\tfrac12 \cdot 0.5 \cdot 2^2 =$$1\ \mathrm{m}$⁠.

The half matters: the speed grows from $0$⁠ to $a t = 1\ \mathrm{m/s}$$a t =$$1\ \mathrm{m/s}$⁠, so the average speed is only $0.5\ \mathrm{m/s}$⁠, and over two seconds that is $1\ \mathrm{m}$⁠. The Scene shows about 0.98 m because of a very small amount of friction, and any answer within $5\%$⁠ is accepted.

Only gravity acts on the ball, and it is vertical. So there is no horizontal force and no horizontal acceleration ($a_x = 0$⁠), and the horizontal part of the velocity does not change from the moment the ball leaves the cannon until it lands: $v_x = v\cos\theta = 20 \cos 45^\circ \approx 14.1\ \mathrm{m/s}$$v_x =$$v\cos\theta =$$20 \cos 45^\circ \approx$$14.1\ \mathrm{m/s}$⁠ for the whole flight.

Only the vertical part changes: it shrinks on the way up, is zero at the top and grows on the way down. You can see this in the Scene: the ghosts are exactly the same distance apart across the field.

Split the motion into two directions, at $60^\circ$⁠ and 20 m/s:

• Sideways there is no force, so $v_x = 20 \cos 60^\circ = 10\ \mathrm{m/s}$$v_x =$$20 \cos 60^\circ =$$10\ \mathrm{m/s}$⁠ the whole time, at the top too.
• Up and down it starts at $v_y = 20 \sin 60^\circ \approx 17.3\ \mathrm{m/s}$$v_y =$$20 \sin 60^\circ \approx$$17.3\ \mathrm{m/s}$⁠, and gravity takes $9.81\ \mathrm{m/s}$⁠ off it every second. The top is the moment when $v_y = 0$⁠.

So at the top the velocity is $10\ \mathrm{m/s}$⁠ sideways, not zero. The acceleration is $g$⁠ downwards throughout, because gravity does not switch off at the top; without it $v_y$⁠ would stay at zero and the ball would never start to fall.

Gravity pulls a ball of mass $m$⁠ with a force $F = mg$⁠, so its acceleration is $a = \dfrac{F}{m} = \dfrac{mg}{m} = g$$a =$$\dfrac{F}{m} =$$\dfrac{mg}{m} =$$g$⁠: the mass cancels. A ball twice as heavy is pulled twice as hard, but it also takes twice the force to accelerate it as much.

The equations of the flight, $x = v\cos\theta\, t$⁠ and $y = v\sin\theta\, t - \tfrac12 g t^2$$y =$$v\sin\theta\, t - \tfrac12 g t^2$⁠, depend only on the speed, the angle and $g$⁠. The same speed and angle give the same path, so the heavier ball lands in the same place.

There is more than one way. At $45^\circ$⁠ the ball goes farthest, so start there and find the speed, or fix a speed and search for the angle.

Without air the range is $R = \dfrac{v^2 \sin 2\theta}{g}$$R =$$\dfrac{v^2 \sin 2\theta}{g}$⁠, and you need $R = 40\ \mathrm{m}$⁠, to within $0.5\ \mathrm{m}$⁠. One way: set Launch speed to 20 m/s and work out the angle:

$\sin 2\theta = \frac{R g}{v^2} = \frac{40 \cdot 9.81}{20^2} \approx 0.981$$\sin 2\theta =$$\frac{R g}{v^2} =$$\frac{40 \cdot 9.81}{20^2} \approx$$0.981$⁠

so $2\theta \approx 78.8^\circ$$2\theta \approx$$78.8^\circ$⁠ and $\theta \approx 39^\circ$$\theta \approx$$39^\circ$⁠. At $39^\circ$⁠ the ball lands about 39.9 m from the line, on the flag. $51^\circ$⁠ works too, because $\sin 2\theta$⁠ is the same for two angles that add up to $90^\circ$⁠. At $45^\circ$⁠ and 20 m/s the ball overshoots, to about 40.8 m.

At the same speed the range depends on the angle only through $\sin 2\theta$⁠, in $R = \dfrac{v^2 \sin 2\theta}{g}$$R =$$\dfrac{v^2 \sin 2\theta}{g}$⁠. Compare the four options:

• $30^\circ$⁠: $\sin 60^\circ \approx 0.87$$\sin 60^\circ \approx$$0.87$⁠
• $45^\circ$⁠: $\sin 90^\circ = 1$⁠
• $60^\circ$⁠: $\sin 120^\circ \approx 0.87$$\sin 120^\circ \approx$$0.87$⁠
• $75^\circ$⁠: $\sin 150^\circ = 0.5$$\sin 150^\circ =$$0.5$⁠

The largest is $1$⁠, at $45^\circ$⁠. At 20 m/s that is about 40.8 m, against about 35.3 m at $30^\circ$⁠ and $60^\circ$⁠ and about 20.4 m at $75^\circ$⁠.

Split the velocity: $v_x = v\cos\theta$⁠ and $v_y = v\sin\theta$⁠. The flight time is $t = \dfrac{2 v_y}{g}$$t =$$\dfrac{2 v_y}{g}$⁠, and the range is $R = v_x t$⁠.

1. Split the velocity: $v_x = 20 \cos 35^\circ \approx 16.38\ \mathrm{m/s}$$v_x =$$20 \cos 35^\circ \approx$$16.38\ \mathrm{m/s}$⁠ and $v_y = 20 \sin 35^\circ \approx 11.47\ \mathrm{m/s}$$v_y =$$20 \sin 35^\circ \approx$$11.47\ \mathrm{m/s}$⁠.
2. The time of flight, until the ball is back at launch height: $t = \dfrac{2 v_y}{g} = \dfrac{2 \cdot 11.47}{9.81} \approx 2.34\ \mathrm{s}$$t =$$\dfrac{2 v_y}{g} =$$\dfrac{2 \cdot 11.47}{9.81} \approx$$2.34\ \mathrm{s}$⁠.
3. The range: $R = v_x t \approx 16.38 \cdot 2.34 \approx 38.3\ \mathrm{m}$$R =$$v_x t \approx$$16.38 \cdot 2.34 \approx$$38.3\ \mathrm{m}$⁠.

Or in one go: $R = \dfrac{v^2 \sin 2\theta}{g} = \dfrac{400 \cdot \sin 70^\circ}{9.81} \approx 38.3\ \mathrm{m}$$R =$$\dfrac{v^2 \sin 2\theta}{g} =$$\dfrac{400 \cdot \sin 70^\circ}{9.81} \approx$$38.3\ \mathrm{m}$⁠. If you got about 26.8 m or 54.7 m, you used the same function, $\sin$⁠ or $\cos$⁠, for both components.

In $R = \dfrac{v^2 \sin 2\theta}{g}$$R =$$\dfrac{v^2 \sin 2\theta}{g}$⁠ the angle does not change, so $R$⁠ is proportional to $v^2$⁠. Doubling $v$⁠ multiplies the range by $2^2 = 4$⁠.

You can see it without the formula too: the time of flight $t = \dfrac{2v\sin\theta}{g}$$t =$$\dfrac{2v\sin\theta}{g}$⁠ doubles, and the horizontal speed $v\cos\theta$⁠ doubles, so the distance $v_x t$⁠ grows four times. In the Scene, at $45^\circ$⁠: 10 m/s gives about 10.2 m, and 20 m/s gives about 40.8 m.

Take the top of the tower as $y = 0$⁠, with up as positive. The field is at $y = -20\ \mathrm{m}$⁠: solve $-20 = v\sin\theta\, t - \tfrac12 g t^2$$-20 =$$v\sin\theta\, t - \tfrac12 g t^2$⁠ and take the positive solution.

Take the top of the tower as $y = 0$⁠, with up as positive. The field is at $y = -20\ \mathrm{m}$⁠.

1. The vertical part of the velocity: $v_y = 10 \sin 30^\circ = 5\ \mathrm{m/s}$$v_y =$$10 \sin 30^\circ =$$5\ \mathrm{m/s}$⁠.
2. The height equation $y = v_y t - \tfrac12 g t^2$$y =$$v_y t - \tfrac12 g t^2$⁠ gives $-20 = 5t - 4.905\, t^2$$-20 =$$5t - 4.905\, t^2$⁠, that is $4.905\, t^2 - 5t - 20 = 0$$4.905\, t^2 - 5t - 20 =$$0$⁠.
3. The quadratic formula: $t = \dfrac{5 \pm \sqrt{5^2 + 4 \cdot 4.905 \cdot 20}}{2 \cdot 4.905} = \dfrac{5 \pm 20.43}{9.81}$$t =$$\dfrac{5 \pm \sqrt{5^2 + 4 \cdot 4.905 \cdot 20}}{2 \cdot 4.905} =$$\dfrac{5 \pm 20.43}{9.81}$⁠.
4. Only the positive solution is a real time: $t \approx 2.59\ \mathrm{s}$$t \approx$$2.59\ \mathrm{s}$⁠.

If you got about 1.02 s, you worked out the time for the ball to come back to the height of the tower, $t = \dfrac{2 v_y}{g}$$t =$$\dfrac{2 v_y}{g}$⁠, without the 20 m drop below it.

Work out the range at two angles, from a 20 m tower at 15 m/s. At each angle, solve $0 = 20 + v_y t - 4.905\, t^2$$0 =$$20 + v_y t - 4.905\, t^2$⁠ for the time and multiply by the horizontal speed:

• At $45^\circ$⁠: $v_x = v_y \approx 10.61\ \mathrm{m/s}$$v_x =$$v_y \approx$$10.61\ \mathrm{m/s}$⁠ and $t \approx 3.37\ \mathrm{s}$$t \approx$$3.37\ \mathrm{s}$⁠, so $R \approx 35.8\ \mathrm{m}$$R \approx$$35.8\ \mathrm{m}$⁠.
• At $30^\circ$⁠: $v_x \approx 12.99\ \mathrm{m/s}$$v_x \approx$$12.99\ \mathrm{m/s}$⁠, $v_y = 7.5\ \mathrm{m/s}$$v_y =$$7.5\ \mathrm{m/s}$⁠ and $t \approx 2.92\ \mathrm{s}$$t \approx$$2.92\ \mathrm{s}$⁠, so $R \approx 38.0\ \mathrm{m}$$R \approx$$38.0\ \mathrm{m}$⁠.

The lower angle goes farther. The drop from the tower already gives the ball time in the air, so it pays to put more of the speed into the horizontal part. $45^\circ$⁠ is best only when the ball lands at its launch height; here the best angle is about $31^\circ$⁠.

Without air, at $45^\circ$⁠ and 25 m/s, the range is $R = \dfrac{v^2}{g} = \dfrac{625}{9.81} \approx 63.7\ \mathrm{m}$$R =$$\dfrac{v^2}{g} =$$\dfrac{625}{9.81} \approx$$63.7\ \mathrm{m}$⁠: that is the dashed curve.

With air a second force acts on the ball, drag, which always points against the direction of motion. Its horizontal part slows the ball sideways, and its vertical part adds to gravity on the way up and cuts the climb short. Both shorten the flight, so the ball lands before the curve, at about 34 m in the Scene. No force can carry it past the curve, because drag only takes speed away.

List the forces on the ball in flight: gravity $mg$⁠ downwards, and drag against the velocity. While the ball climbs forwards at a slant, drag points backwards and downwards, so it has two parts:

• a horizontal part backwards, which reduces $v_x$⁠. Without air $v_x$⁠ would stay constant.
• a vertical part downwards on the way up, which adds to gravity and shortens the climb.

The ball moves more slowly sideways and spends less time in the air, so it lands closer. Its weight and the pull of gravity do not change: the only new thing is drag.

Without air, $45^\circ$⁠ balances time in the air against horizontal speed. With a lot of drag every moment in the air has a cost: drag takes speed from the ball all the time, and a ball that flies high spends most of its speed going up and coming down. A lower path spends less time in the air and keeps more of its horizontal speed, so it goes farther.

In the Scene, for a shuttlecock at 25 m/s: at $45^\circ$⁠ it reaches about 8.5 m, and at $32^\circ$⁠ about 9.1 m, the longest range. Try it yourself with Launch angle.