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.

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 (AQA A Level Maths: Mechanics): Flashcards

Exam code: 7357

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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 (11)

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.
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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