In mechanics, what is a projectile ?

A projectile is a particle moving freely under gravity in two dimensions. Moving freely means that no force acts on it except its own weight. In particular nothing acts on it horizontally at any point during the motion.

A stone is thrown from the top of a cliff rather than from ground level. Which of the two directions does the cliff height affect, and which does it not?

It affects the vertical direction only. The horizontal motion is unchanged, because the horizontal acceleration is zero wherever the stone started from, so the horizontal equation is the same as it would be on level ground. The cliff enters the problem only through the vertical displacement at the moment of landing, which is measured from the point of projection rather than from the ground.

A projectile is launched and lands at the same height . What does the symmetry of its path let you say about the highest point?

The path is symmetrical about the highest point, so the projectile reaches it exactly halfway through the motion. The time to the highest point is therefore half the total time of flight, and the horizontal distance to it is half the range. Both of these depend on the launch and landing heights being equal: thrown from a cliff, the path is no longer symmetrical and neither shortcut holds.

What three assumptions do all the standard projectile formulae depend on?

They assume that the launch and landing points are at the same vertical level , that the projectile travels over horizontal ground , and that no force acts on it except gravity . The first is the one that is most often broken by a question. Anything launched from a cliff, a building or a window lands lower than it started, and the formulae no longer apply.

True or False? You can assume the range formula gives the horizontal distance travelled by a projectile launched from the top of a cliff.

False. Every one of the standard projectile formulae is derived on the assumption that the launch and landing points are at the same level , and a cliff breaks that. A projectile launched from a cliff is in the air for longer than the formula allows for, so it travels further horizontally than the formula predicts. For a launch and landing at different heights, go back to the constant-acceleration formulae applied to each direction separately.

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Projectiles (Edexcel A Level Maths: Mechanics): Flashcards

Exam code: 9MA0

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FrontHorizontal & Vertical Components of Projectiles
In mechanics, what is a projectile?
BackHorizontal & Vertical Components of Projectiles
No force acts on it horizontally, so the horizontal component of the acceleration is zero.
The consequence is that the horizontal component of the velocity never changes: it is the same at launch, at the highest point and at landing.
Taking upwards as positive, the whole acceleration is $a = 0 i - g j$ , so the only acceleration anywhere in the motion is the vertical one due to gravity.

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Cards in this collection (22)

In mechanics, what is a projectile?
A projectile is a particle moving freely under gravity in two dimensions.
Moving freely means that no force acts on it except its own weight. In particular nothing acts on it horizontally at any point during the motion.
Why does a projectile have no horizontal acceleration, and what does that mean for its motion?
No force acts on it horizontally, so the horizontal component of the acceleration is zero.
The consequence is that the horizontal component of the velocity never changes: it is the same at launch, at the highest point and at landing.
Taking upwards as positive, the whole acceleration is $a = 0 i - g j$ , so the only acceleration anywhere in the motion is the vertical one due to gravity.
A projectile is launched with speed $U$ at an angle $θ$ to the horizontal. Fill in the blanks with the components of its initial velocity:
horizontal component $u_{x} =______$
vertical component $u_{y} =______$
The completed components are:
horizontal component $u_{x} = U \cos θ$
vertical component $u_{y} = U \sin θ$
The launch velocity is the hypotenuse of a right-angled triangle whose other two sides are the components, so the angle is measured from the horizontal side, which is why the cosine goes with the horizontal.
True or False?
You can assume that the initial vertical component of a projectile's velocity is positive.
False.
A projectile can be launched below the horizontal, for example thrown downwards from a window or fired from a turret at something on the ground.
The angle $θ$ is then taken as negative, which makes $\sin θ$ negative and so makes $u_{y} = U \sin θ$ negative, pointing downwards.
The horizontal component is unaffected, because $\cos θ$ is the same for $θ$ and $- θ$ .
A projectile is launched with speed $U$ at an angle $θ$ . What is the difference between $U$ and $u$ ?
$U$ is the initial speed, a single number with no direction.
$u$ is the initial velocity, a vector with components $U \cos θ$ and $U \sin θ$ .
So $U$ is the magnitude of $u$ , and keeping them apart matters because the components carry the angle and $U$ does not.
How are the constant-acceleration formulae applied to a projectile?
Apply them to the horizontal and vertical directions separately, as two one-dimensional problems.
The acceleration is different in each: it is zero horizontally, which reduces the horizontal formula to $s = u t$ , and it is $- g$ vertically, taking upwards as positive.
Because each direction is treated on its own, $v^{2} = u^{2} + 2 a s$ can be used here, even though it has no vector form.
What single quantity links the horizontal and vertical parts of a projectile problem?
The time, $t$ .
The two directions are otherwise independent, but the projectile is in flight for one interval of time, so the same $t$ appears in both sets of equations.
That is what makes the problems solvable: a time found from the vertical motion can be substituted into the horizontal motion, and the other way round.
True or False?
You can assume that a projectile is momentarily at rest at its highest point.
False.
Only the vertical component of the velocity is zero there; the horizontal component is unchanged, because there is no horizontal acceleration.
So at the highest point the projectile is still moving horizontally, with speed $U \cos θ$ , which is the smallest speed it has at any point in its flight and is never zero.
An object thrown straight up is the special case that does come to rest, because it has no horizontal component to begin with.
How do you find the greatest height a projectile reaches?
Use the fact that the vertical component of the velocity is zero at that point, and apply a vertical constant-acceleration formula.
With $v = 0$ , $u = U \sin θ$ and $a = - g$ , the formula $v^{2} = u^{2} + 2 a s$ gives the vertical displacement directly, without needing the time.
If the projectile was launched from above ground level, that displacement is measured from the launch point, so the launch height has to be added to get the height above the ground.
A stone is thrown from the top of a cliff rather than from ground level. Which of the two directions does the cliff height affect, and which does it not?
It affects the vertical direction only.
The horizontal motion is unchanged, because the horizontal acceleration is zero wherever the stone started from, so the horizontal equation is the same as it would be on level ground.
The cliff enters the problem only through the vertical displacement at the moment of landing, which is measured from the point of projection rather than from the ground.
A projectile is launched and lands at the same height. What does the symmetry of its path let you say about the highest point?
The path is symmetrical about the highest point, so the projectile reaches it exactly halfway through the motion.
The time to the highest point is therefore half the total time of flight, and the horizontal distance to it is half the range.
Both of these depend on the launch and landing heights being equal: thrown from a cliff, the path is no longer symmetrical and neither shortcut holds.
The constant-acceleration formulae give the horizontal and vertical displacements of a projectile separately. How is the equation of its trajectory different?
The two suvat results are parametric equations: they give $x$ and $y$ each in terms of the time, which acts as the parameter.
The equation of the trajectory is a Cartesian equation, linking $x$ and $y$ to each other directly, in the form $y = f (x)$ , with no time in it at all.
It is obtained by eliminating the parameter between the two, exactly as for any pair of parametric equations.
When deriving the equation of a trajectory, which of the two displacement equations do you rearrange, and why that one?
Rearrange the horizontal equation.
Because there is no horizontal acceleration it is simply $x = (U \cos θ) t$ , which rearranges to $t = \frac{x}{U \cos θ}$ in one step.
The vertical equation contains $t^{2}$ as well as $t$ , so making $t$ its subject would mean solving a quadratic; substituting the horizontal result into the vertical one avoids that entirely.
How does the equation of a trajectory show that the path of a projectile is a parabola?
The equation is $y = x \tan θ - \frac{g x^{2}}{2 U^{2} \cos^{2} θ}$ .
For a fixed launch speed and angle, everything except $x$ and $y$ is constant, so this is a quadratic in $x$ , and the graph of a quadratic is a parabola.
The coefficient of $x^{2}$ is negative, since $g$ , $U^{2}$ and $\cos^{2} θ$ are all positive, so the parabola opens downwards, which is why the path has a greatest height rather than a least one.
A projectile is launched with speed $U$ at an angle $θ$ . Fill in the blanks in the equation of its trajectory:
$y = x______- \frac{g x^{2}}{2 U^{2}______}$
The completed equation is:
$y = x \tan θ - \frac{g x^{2}}{2 U^{2} \cos^{2} θ}$
The tangent comes from $\frac{U \sin θ}{U \cos θ}$ in the first term, and the $\cos^{2} θ$ from squaring the $U \cos θ$ in the denominator of the substituted time.
What can you do with the equation of a trajectory that you cannot do as directly with the two separate displacement equations?
Get straight from a horizontal distance to a height, or from a height to a horizontal distance, without finding the time first.
With the parametric pair you would have to find $t$ from one equation and substitute it into the other every time; the trajectory equation has already done that once and for all.
That is what makes it the natural tool for questions asking whether a projectile clears a wall or passes over a given point.
What three assumptions do all the standard projectile formulae depend on?
They assume that the launch and landing points are at the same vertical level, that the projectile travels over horizontal ground, and that no force acts on it except gravity.
The first is the one that is most often broken by a question. Anything launched from a cliff, a building or a window lands lower than it started, and the formulae no longer apply.
How is the time of flight of a projectile derived, and why does the working produce two answers?
Set the vertical displacement to zero, since the projectile returns to the level it started from, and use $s = u t + \frac{1}{2} a t^{2}$ vertically:
$0 = (U \sin θ) t - \frac{1}{2} g t^{2}$
Factorising gives $0 = t (U \sin θ - \frac{1}{2} g t)$ , so either $t = 0$ or $t = \frac{2 U \sin θ}{g}$ .
Two answers appear because the vertical displacement is zero at two moments, at the start and at landing; the $t = 0$ root is the launch itself and is rejected.
At the maximum height the vertical component of velocity is zero. Which vertical formula gives the time to that point, and which gives the height itself?
For the time, use $v = u + a t$ : with $v = 0$ , $0 = U \sin θ - g t$ , giving $t = \frac{U \sin θ}{g}$ .
For the height, use $v^{2} = u^{2} + 2 a s$ : with $v = 0$ , $0 = U^{2} \sin^{2} θ - 2 g y$ , giving $y = \frac{U^{2} \sin^{2} θ}{2 g}$ .
Choosing between them is the usual suvat choice: the first formula contains $t$ and not $s$ , the second contains $s$ and not $t$ .
For a projectile launched with speed $U$ at an angle $θ$ over horizontal ground, fill in the blanks:
time of flight $= \frac{2 U______}{g}$
range $= \frac{U^{2}______}{g}$
The completed formulae are:
time of flight $= \frac{2 U \sin θ}{g}$
range $= \frac{U^{2} \sin 2 θ}{g}$
Note the difference: the time of flight has $\sin θ$ , the range has $\sin 2 θ$ . The doubled angle appears only in the range, and only because of the identity used at the last step of its derivation.
How is the range of a projectile derived, and where does the $\sin 2 θ$ come from?
The range is the horizontal distance covered in the whole time of flight, and horizontally there is no acceleration, so $s = u t$ applies:
$x = (U \cos θ) (\frac{2 U \sin θ}{g}) = \frac{2 U^{2} \sin θ \cos θ}{g}$
The double angle identity $2 \sin θ \cos θ = \sin 2 θ$ then turns that into $\frac{U^{2} \sin 2 θ}{g}$ .
So the $\sin 2 θ$ is not a mechanics result at all: it is a trigonometric identity tidying up the answer at the very end.
True or False?
You can assume the range formula gives the horizontal distance travelled by a projectile launched from the top of a cliff.
False.
Every one of the standard projectile formulae is derived on the assumption that the launch and landing points are at the same level, and a cliff breaks that.
A projectile launched from a cliff is in the air for longer than the formula allows for, so it travels further horizontally than the formula predicts.
For a launch and landing at different heights, go back to the constant-acceleration formulae applied to each direction separately.

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