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Slopes, moving platforms and the farthest reach

A ball thrown up a hill, a ball thrown from a moving lorry, and the farthest point that a throw can reach. Tilted axes make the hill simple, a change of frame makes the lorry simple, and the envelope of all the paths shows the limit.

This is the first lesson of Slopes, Chases and Changing Speed: Plane Motion Pushed Further, free for everyone to read.

A throw on a hillside

A golfer stands on a fairway that rises at a steady slope and hits the ball up the hill. Does the level-ground formula $R = \dfrac{u^2\sin 2\theta}{g}$ still work? No. That formula is for a ball that lands at the height where it started. On a slope the ground rises under the ball.

The way out is to tilt the axes. Take one axis along the slope and one at right angles to the slope. The ball is the same ball and gravity is the same gravity. Only the directions of the axes have changed. Gravity, which points straight down, splits into two parts:

$g\sin\alpha$ along the slope, pointing down the hill,
$g\cos\alpha$ at right angles to the slope, pointing into the hill.

Here $\alpha$ is the angle of the hillside with the horizontal. Let the ball leave at speed $u$ at an angle $\beta$ above the slope. Then the angle above the horizontal is $\theta = \alpha + \beta$ .

Direction	Starting speed	Acceleration
Along the slope	$u\cos\beta$	$-g\sin\alpha$
At right angles	$u\sin\beta$	$-g\cos\alpha$

A hillside that rises to the right at an angle alpha to the horizontal, drawn as a grey line with a white ground below it. A ball leaves a point on the hillside along a blue curved path and lands farther up the hillside. Two tilted axes are drawn at the launch point: one along the hillside, marked x prime, and one at right angles to it, marked y prime. A blue arrow shows the launch velocity u, at an angle beta above the hillside. A red arrow pointing straight down from the launch point is the acceleration g. Two shorter arrows show its parts: g sine alpha along the hillside, pointing down the hill, and g cosine alpha at right angles to the hillside, pointing into the hill.

The ball lands when it comes back to the slope. That happens when the motion at right angles to the slope is finished. From $0 = u\sin\beta\,T - \tfrac{1}{2}g\cos\alpha\,T^2$ we get

\displaystyle T = \frac{2u\sin\beta}{g\cos\alpha}

During this time the ball moves along the slope. The distance along the slope is the range $R$ :

\displaystyle R = u\cos\beta\,T - \tfrac{1}{2}g\sin\alpha\,T^2 = \frac{2u^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}

The second form uses $\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ . The greatest distance of the ball from the slope is

\displaystyle H_{\perp} = \frac{(u\sin\beta)^2}{2g\cos\alpha}

Put $\alpha = 0$ and you get the level-ground formulas again. That is a good check on any formula for a slope.

A worked example. A ball leaves a hillside of $20^\circ$ at $24\ \text{m/s}$ , at $45^\circ$ above the slope. Find the time in the air, the range along the slope and the greatest distance from the slope.

The time. $T = \dfrac{2 \times 24\ \text{m/s} \times \sin 45^\circ}{9.8\ \text{m/s}^2 \times \cos 20^\circ} = 3.69\ \text{s}$ .
The range. The numerator is $2 \times (24\ \text{m/s})^2 \times \sin 45^\circ \times \cos 65^\circ = 344.3\ \text{m}^2/\text{s}^2$ . The denominator is $9.8\ \text{m/s}^2 \times \cos^2 20^\circ = 8.65\ \text{m/s}^2$ . So $R = 39.78\ \text{m}$ .
The greatest distance from the slope. $H_{\perp} = \dfrac{(24\ \text{m/s} \times \sin 45^\circ)^2}{2 \times 9.8\ \text{m/s}^2 \times \cos 20^\circ} = 15.64\ \text{m}$ .

Check. Use ordinary horizontal and vertical axes. The ball is at $x = u\cos\theta\,T = 37.38\ \text{m}$ and $y = u\sin\theta\,T - \tfrac{1}{2}gT^2 = 13.61\ \text{m}$ . The ratio is $y/x = 0.364$ , which is $\tan 20^\circ = 0.364$ . So the ball is on the slope. The distance along the slope is $\sqrt{x^2 + y^2} = 39.78\ \text{m}$ . It fits.

Set the slope to $0^\circ$ and the angle above the slope to $45^\circ$ . Which range do you get? Now raise the slope. How does the farthest range change?

The farthest throw up and down a slope

For a fixed speed, which angle gives the longest range along the slope? The identity $2\sin A\cos B = \sin(A+B) + \sin(A-B)$ turns the range into $R = \dfrac{u^2}{g\cos^2\alpha}\left[\sin(2\beta + \alpha) - \sin\alpha\right]$ . Ask when the first sine is largest. It is largest when $2\beta + \alpha = 90^\circ$ . So

\displaystyle \beta = 45^\circ - \frac{\alpha}{2} \qquad R_{\max} = \frac{u^2}{g(1+\sin\alpha)} \quad \text{(up the slope)}

The ball leaves midway between the slope and the vertical. For a slope that falls away from the thrower, put $-\alpha$ in place of $\alpha$ in every formula above. The same working gives

\displaystyle \beta = 45^\circ + \frac{\alpha}{2} \qquad R_{\max} = \frac{u^2}{g(1-\sin\alpha)} \quad \text{(down the slope)}

For $u = 24\ \text{m/s}$ and $\alpha = 20^\circ$ , the farthest throw up the hill is $43.8\ \text{m}$ , at $\beta = 35^\circ$ . Down the hill it is $89.3\ \text{m}$ , at $\beta = 55^\circ$ . On level ground it would be $58.8\ \text{m}$ . Does that make sense? Down the hill the ground drops away, so the ball keeps falling for longer. Up the hill the ground comes up to meet it.

Two more results are worth knowing.

The ball strikes the slope at right angles (going up). The speed along the slope must be zero at the landing: $u\cos\beta - g\sin\alpha\,T = 0$ . This gives $\tan\beta\,\tan\alpha = \tfrac{1}{2}$ . For $\alpha = 30^\circ$ the angle is $\beta = 40.9^\circ$ .
A ball thrown horizontally from a point on a slope that falls away. Here the ball leaves at $\beta = \alpha$ above the surface of the slope. The time is $T = \dfrac{2u\tan\alpha}{g}$ and the distance along the slope is $\dfrac{2u^2\sin\alpha}{g\cos^2\alpha}$ . For $u = 10\ \text{m/s}$ and $\alpha = 30^\circ$ these are $1.18\ \text{s}$ and $13.61\ \text{m}$ .

A throw from a moving platform

A boy on an open lorry moves along a straight road at a steady speed $v$ . He throws a ball. What does he see, and what does a girl standing by the road see?

The rule. The velocity of the ball over the road is the velocity of the lorry plus the velocity of the ball relative to the lorry:

\displaystyle \vec{v}_{\text{ball, road}} = \vec{v}_{\text{lorry}} + \vec{v}_{\text{ball, lorry}}

Gravity acts in the same way in both frames when the lorry moves at a steady velocity. So the up and down motion is the same for both watchers. Only the horizontal speed differs. If the boy throws at speed $u$ , at an angle $\varphi$ from the forward direction and above the horizontal, then

the time in the air is $T = \dfrac{2u\sin\varphi}{g}$ for both,
the boy sees the ball cover $u\cos\varphi\,T$ along the road,
the girl sees it cover $(v + u\cos\varphi)\,T$ .

A negative $\cos\varphi$ means a throw towards the back of the lorry.

A worked example. A lorry moves at $12\ \text{m/s}$ . The boy throws a ball backwards at $16\ \text{m/s}$ relative to the lorry. At what angle must he throw so that the girl sees it go straight up and down?

The condition. The ball must have no horizontal speed over the road. So $v + u\cos\varphi = 0$ , which gives $\cos\varphi = -\dfrac{12\ \text{m/s}}{16\ \text{m/s}}$ .
The angle. The throw is $41.4^\circ$ above the horizontal measured from the backward direction. The vertical speed is $16\ \text{m/s} \times \sin 41.4^\circ = 10.58\ \text{m/s}$ .
The time and the height. $T = \dfrac{2 \times 10.58\ \text{m/s}}{9.8\ \text{m/s}^2} = 2.16\ \text{s}$ and $H = \dfrac{(10.58\ \text{m/s})^2}{2 \times 9.8\ \text{m/s}^2} = 5.71\ \text{m}$ .
What the boy sees. The lorry moves $12\ \text{m/s}$ $\times$ $2.16\ \text{s}$ $= 25.9\ \text{m}$ while the ball is up. So the ball lands $25.9\ \text{m}$ behind him.

Two views of a ball thrown backwards from a lorry. The lorry is a box on wheels that moves to the right with a speed v. In the first view, seen from the road, the ball goes straight up and straight down, drawn as a vertical blue line above a point on the road, while the lorry moves on to the right. In the second view, seen from the lorry, the ball follows a curved path to the left of the thrower and lands behind the lorry. A grey double arrow shows the distance that the lorry moves while the ball is in the air.

Now suppose the lorry speeds up at a steady $a$ . Once the ball has left the hand, no force acts on it along the road. So it keeps its horizontal speed, while the lorry gains speed. For the boy, the ball seems to fall behind with the acceleration $a$ towards the back. The up and down motion does not change, because the lorry accelerates only along the road.

A second example. The lorry speeds up at $3.0\ \text{m/s}^2$ . The boy throws a ball straight up at $8\ \text{m/s}$ relative to the lorry. The time in the air is $T = \dfrac{2u}{g} = 1.63\ \text{s}$ . The lorry gains $\tfrac{1}{2}aT^2 = 4.00\ \text{m}$ on the ball. The ball lands $4.00\ \text{m}$ behind the boy.

Put the speed of the lorry at $12\ \text{m/s}$ and the speed of the ball at $16\ \text{m/s}$ . Change the angle until the landing point and the point under the throw come together. Which angle is it, and why?

The farthest reach and the envelope

So far we asked for one angle. Now we ask a bigger question: which points can a ball thrown at speed $u$ reach at all? The path of a ball thrown at angle $\theta$ is

\displaystyle y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta} = x\tan\theta - \frac{g x^2}{2u^2}\left(1 + \tan^2\theta\right)

Pick a point $(x, y)$ . The equation is a quadratic in $\tan\theta$ . It can have two solutions (two angles reach the point), one solution (a single angle), or none (the point is out of reach). The boundary between "two" and "none" is where the two solutions meet. Put the discriminant equal to zero:

\displaystyle y = \frac{u^2}{2g} - \frac{g x^2}{2u^2}

This curve is the envelope of all the paths of speed $u$ . Every possible path lies below it and touches it once. (Some books call it the bounding parabola or the safety parabola.) Points on the curve can be reached with one angle. Points under the curve can be reached with two angles. Points above it cannot be reached at all.

A graph of height against horizontal distance for a launch speed of 20 metres per second. Six thin blue parabolic paths leave the origin at angles of 15, 30, 45, 60, 75 and 85 degrees, and each one stays below a thick amber curve. The amber curve is an upside-down parabola with its top, 20.4 metres high, directly above the origin. It crosses the ground at 40.8 metres on each side of the origin. Each blue path touches the amber curve at one point. The amber curve is labelled the envelope.

The envelope answers many questions at once.

Level ground. Put $y = 0$ : $x = \dfrac{u^2}{g}$ . This is the farthest range. For $u = 20\ \text{m/s}$ it is $40.82\ \text{m}$ , and the top of the envelope is $20.41\ \text{m}$ high.
A slope. The farthest point on a slope is where the line of the slope meets the envelope. This gives the formula $\dfrac{u^2}{g(1+\sin\alpha)}$ that we found with tilted axes.
The least speed to reach a point. A point $(x, y)$ lies on the envelope for the least speed that can reach it. Then $u^4 - 2gy\,u^2 - g^2x^2 = 0$ , so

\displaystyle u_{\min}^2 = g\left(y + \sqrt{x^2 + y^2}\right) \qquad \tan\theta = \frac{y + \sqrt{x^2 + y^2}}{x}

A window. A window is $20\ \text{m}$ away horizontally and $12\ \text{m}$ above your hand. The distance to it is $\sqrt{20^2 + 12^2}\ \text{m} = 23.32\ \text{m}$ . So $u_{\min}^2 = 9.8\ \text{m/s}^2 \times (12\ \text{m} + 23.32\ \text{m}) = 346.2\ \text{m}^2/\text{s}^2$ and $u_{\min} = 18.61\ \text{m/s}$ . The angle is $\theta = 60.5^\circ$ , steeper than $45^\circ$ because the window is high.

Throwing from a height

A cliff changes the picture. A stone thrown from a height $h$ lands on level ground below. The farthest throw is not at $45^\circ$ . The farthest landing point is where the envelope reaches the ground, at $y = -h$ . Putting $y = -h$ into the envelope and into the quadratic gives

\displaystyle \tan\theta = \frac{u}{\sqrt{u^2 + 2gh}} \qquad R_{\max} = \frac{u}{g}\sqrt{u^2 + 2gh}

A worked example. A stone is thrown at $18\ \text{m/s}$ from the top of a cliff $25\ \text{m}$ high.

The square root. The expression under the root is $u^2 + 2gh = (18\ \text{m/s})^2 + 2 \times 9.8\ \text{m/s}^2 \times 25\ \text{m} = 814\ \text{m}^2/\text{s}^2$ . So $\sqrt{u^2 + 2gh} = 28.53\ \text{m/s}$ .
The best angle. $\tan\theta = \dfrac{18}{28.53}$ , so $\theta = 32.2^\circ$ .
The longest range. $R_{\max} = \dfrac{18\ \text{m/s}}{9.8\ \text{m/s}^2} \times 28.53\ \text{m/s} = 52.40\ \text{m}$ .

A throw at $45^\circ$ goes only $49.69\ \text{m}$ , and one at $30^\circ$ goes $52.33\ \text{m}$ . The curve of range against angle is very flat near its top. Missing the best angle by a few degrees costs little.

Practise with new numbers

Try it in Python

The program throws a ball at $20\ \text{m/s}$ from the foot of a hillside of $25^\circ$ , up the hill and down the hill. It tries every angle from $1^\circ$ to $89^\circ$ above the slope, in steps of $0.1^\circ$ , and keeps the farthest. Then it compares the result with the formula.

Common mistakes

Using $\dfrac{u^2\sin 2\theta}{g}$ on a slope. It is for a landing at the height of the launch. Use tilted axes.
Taking the angle of the throw from the horizontal in the slope formulas. In $T = \dfrac{2u\sin\beta}{g\cos\alpha}$ the angle $\beta$ is measured from the slope.
Forgetting that the ball keeps the speed of the lorry. Seen from the road, a ball thrown backwards at the speed of the lorry goes straight up and down.
Saying that the best angle is always $45^\circ$ . It is $45^\circ$ only when the launch and the landing are at the same height on level ground.
Reading the envelope as a path. No ball moves along the envelope. It is the edge of the region that all the paths together cover.
Mixing the frames. Choose the frame of the road or the frame of the lorry. Do not take the speed from one and the distance from the other.

Remember these

On a slope, take axes along the slope and at right angles to it. The accelerations are $-g\sin\alpha$ and $-g\cos\alpha$ .
$T = \dfrac{2u\sin\beta}{g\cos\alpha}$ and $R = \dfrac{2u^2\sin\beta\cos(\alpha+\beta)}{g\cos^2\alpha}$ , with $\beta$ measured from the slope.
Farthest range: $\dfrac{u^2}{g(1+\sin\alpha)}$ up a slope, $\dfrac{u^2}{g(1-\sin\alpha)}$ down it, at $\beta = 45^\circ \mp \dfrac{\alpha}{2}$ .
From a platform that moves at a steady velocity, add the velocity of the platform. The up and down motion does not change.
The envelope is $y = \dfrac{u^2}{2g} - \dfrac{gx^2}{2u^2}$ . The least speed to reach $(x, y)$ is $u_{\min}^2 = g\left(y + \sqrt{x^2+y^2}\right)$ .
From a height $h$ to level ground, $R_{\max} = \dfrac{u}{g}\sqrt{u^2 + 2gh}$ at $\tan\theta = \dfrac{u}{\sqrt{u^2+2gh}}$ .

References

OpenStax, University Physics Volume 1, Chapter 4, Motion in Two and Three Dimensions (projectile motion).
D. Halliday, R. Resnick and J. Walker, Fundamentals of Physics, Chapter 4 (projectile motion).

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

A ball is thrown up a slope of $30^\circ$ . What are the accelerations along the slope and at right angles to it?

A girl throws a ball straight up inside a lorry that moves at a steady $20\ \text{m/s}$ . Where does the ball land? What path does a boy by the road see?

A ball leaves at $18\ \text{m/s}$ with the best angle for each case. How far does it go along a slope of $20^\circ$ , up the hill and down the hill?

A target is on the ground, $30\ \text{m}$ away from a thrower on level ground. What is the least speed that can reach it?

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Slopes, Chases and Changing Speed: Plane Motion Pushed Further is a Standard course. It adds 2 more lessons, practice and tests on top of the free chapter.

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