Projectile Motion

Projectile Motion
9:32 — by lasseviren1

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

• Projectile motion is parabolic because the vertical position of the object is influenced only by a constant acceleration, (if constant drag etc. is also assumed) and also because horizontal velocity is generally constant.

Put simply, basic projectile motion is parabolic because its related equation of motion,

$x \left(t\right) = \frac{1}{2} a {t}^{2} + {v}_{i} t + {x}_{i}$

is quadratic, and therefore describes a parabola.

However, I can explain a bit more in-depth why this works, if you'd like, by doing a little integration. Starting with a constant acceleration,

$a = k$,

we can move on to velocity by integrating with respect to $t$. ($a = k$ is interpreted as being $a = k {t}^{0}$)

$v \left(t\right) = \int k \mathrm{dt} = k t + {v}_{i}$

The constant of integration here is interpreted to be initial velocity, so I've just named it ${v}_{i}$ instead of $C$.

Now, to position:

$x \left(t\right) = \int \left(k t + {v}_{i}\right) \mathrm{dt}$
$x \left(t\right) = \frac{1}{2} k {t}^{2} + {v}_{i} t + {x}_{i}$

Again, the constant of integration is interpreted in this case to be initial position. (denoted ${x}_{i}$)

Of course, this equation will probably look familiar to you. It's the equation of motion I described above.

Don't worry if you haven't learned about integration yet; the only thing you need to worry about is the power of $t$ as we move from acceleration to velocity to position. If $t$ was present in the initial $a = k$ equation, with a degree other than $0$, (in other words, if $a$ is changing over time) then after integration we would end up with a degree different from $2$. But since $a$ is constant, $t$ will always be squared in the equation for position, resulting in a parabola.

Since acceleration due to gravity is generally fairly constant at around $9.8 \frac{m}{s} ^ 2$, we can say that the trajectory of a projectile is parabolic.

A case where the path wouldn't appear to be parabolic is if an object were dropped, falling straight downwards, with no horizontal velocity. In this case the path looks more like a line, but it's actually a parabola which has been infinitely horizontally compressed. In general, the smaller horizontal velocity, the more the parabola is compressed horizontally.

• In the absence of air resistance there are no forces or components of forces that act horizontally.

A velocity vector can only change if there is acceleration (acceleration is the rate of change of velocity). In order to accelerate a resultant force is required (according to Newton's Second Law, $\vec{F} = m \vec{a}$).

In the absence of air resistance the only force acting on a projectile in flight is the weight of the object. Weight by definition acts vertically downwards, hence no horizontal component.

• Gravity is a very influential force when doing calculations with projectile motion.

When a projectile flies in an empty system, like space, it tends to fly in a straight line because there is nothing to change the trajectory. But when you add gravity, the projectile will now accelerate towards the source of gravity. So when you shoot something, it will accelerate downwards and eventually hit the ground. Without gravity, it would fly in a straight line.

• Actually whilst the ball is in contact with the foot it is not a projectile. A kicked football is an example of a projectile (i.e. after it has been kicked).

A projectile is an object that moves under the influence of gravity, what that means is that it's weight is the only force that acts upon it. In reality there is a drag force too, but that is frequently ignored for the purpose of projectile calculations.

Whilst being kicked the ball has a normal reaction force from the foot acting upon it in addition to its weight. So that does not count as a projectile. After being kicked the ball only has its weight (and drag) acting upon it, so it is a projectile. Whilst in flight the ball will continue with constant horizontal velocity (no horizontal forces) and experience a constant downwards vertical acceleration (due to its weight).

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