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.

True or False? A projectile takes the same time to rise to its highest point as it then takes to fall back to the height it was launched from.

True. The vertical motion has the same constant acceleration throughout, and that stretch of the path starts and finishes at the same height, so the two halves are mirror images of each other. It returns to the launch height with a vertical speed equal to the one it set off with, now directed downwards.

A projectile is launched towards a wall. What has to be compared in order to decide whether it passes over the wall?

The height of the projectile when it has travelled the wall's horizontal distance , compared with the height of the wall . Its greatest height is not the right thing to use, because the projectile may reach that height well before or well after it arrives at the wall.

Define the trajectory of a projectile.

The trajectory is the path that the projectile follows through the air during its motion. Under the standard modelling assumptions that path is a parabola .

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 (24)

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 connects the horizontal and vertical working in a projectile problem?
The time, $t$ .
It is the one quantity shared by both directions, because a single value of $t$ describes the same instant horizontally and vertically.
So the usual route through a problem is to find $t$ from one direction and substitute it into the other.
Why does the horizontal motion of a projectile need only $s = u t$ , rather than one of the fuller suvat formulae?
Because the horizontal acceleration is zero, so every term containing $a$ vanishes.
That turns $s = u t + \frac{1}{2} a t^{2}$ into $s_{x} = u_{x} t$ , which is the only horizontal equation a projectile problem ever needs.
A projectile is launched and lands at the same height. Fill in the missing word:
The horizontal distance from the launch point to the highest point is $______$ of the range.
The horizontal distance from the launch point to the highest point is half of the range.
The path is symmetrical about the highest point, and the projectile covers equal horizontal distances on the way up and on the way down.
What is a projectile's speed at the highest point of its path?
Its speed there is the horizontal component of its launch velocity, which is $U \cos θ$ for a launch at speed $U$ and angle $θ$ .
Only the vertical part of the velocity has been brought to zero by gravity, and the horizontal part is untouched.
So the projectile is still moving at the top, and its speed there is a minimum rather than zero.
True or False?
A projectile takes the same time to rise to its highest point as it then takes to fall back to the height it was launched from.
True.
The vertical motion has the same constant acceleration throughout, and that stretch of the path starts and finishes at the same height, so the two halves are mirror images of each other.
It returns to the launch height with a vertical speed equal to the one it set off with, now directed downwards.
A projectile is launched towards a wall. What has to be compared in order to decide whether it passes over the wall?
The height of the projectile when it has travelled the wall's horizontal distance, compared with the height of the wall.
Its greatest height is not the right thing to use, because the projectile may reach that height well before or well after it arrives at the wall.
Define the trajectory of a projectile.
The trajectory is the path that the projectile follows through the air during its motion.
Under the standard modelling assumptions that path is a parabola.
What is the difference between the parametric and Cartesian forms of a projectile's path?
The parametric form gives $x$ and $y$ as two separate equations, each in terms of the time $t$ .
The Cartesian form is a single equation linking $x$ and $y$ directly, $y = f (x)$ , with no $t$ in it at all.
How do you turn the two projectile equations into a Cartesian equation?
Eliminate $t$ between them.
Rearrange the horizontal equation $x = (U \cos θ) t$ to give $t = \frac{x}{U \cos θ}$ , then substitute that into the vertical equation for $y$ .
The horizontal equation is the one to rearrange because it is the simpler of the two, having no $t^{2}$ term in it.
Complete the general equation for the trajectory of a projectile launched from the origin with speed $U$ at an angle $θ$ to the horizontal:
$y = x \tan θ - \frac{g x^{2}}{______}$
The completed equation is:
$y = x \tan θ - \frac{g x^{2}}{2 U^{2} \cos^{2} θ}$
The launch speed appears squared in the denominator, and the angle appears there as $\cos^{2} θ$ rather than as the $\tan θ$ of the first term.
True or False?
The equation of a projectile's trajectory gives its height at a given time.
False.
It gives the height at a given horizontal distance, $x$ .
Time has been eliminated from it altogether, which is the whole point of putting the path into Cartesian form; for the height at a given time, use the vertical suvat equation instead.
Where does the $\tan θ$ in a projectile's trajectory equation come from?
From the first term after the substitution, $\frac{(U \sin θ) x}{U \cos θ}$ .
The $U$ cancels, leaving $\frac{\sin θ}{\cos θ}$ , which is $\tan θ$ .
Note that $U$ does not cancel in the second term, where it is squared, so the launch speed has not disappeared from the equation.
A projectile is launched at $16 {m s}^{- 1}$ at $70 °$ to the horizontal, with trajectory $y = x \tan θ - \frac{g x^{2}}{2 U^{2} \cos^{2} θ}$ . Find its height after it has travelled 12 m horizontally, to 2 significant figures.
Its height is $9.4 m$ .
Substituting gives $y = 12 \tan 70 ° - \frac{9.8 \times 12^{2}}{2 \times 16^{2} \cos^{2} 70 °} = 32.97 - 23.56 = 9.407 \dots$
The angle must be entered in degrees, and squaring the cosine before multiplying keeps the denominator right.
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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