Projectile Motion

Key Questions

• 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 opposes the vertical component of velocity of the projectile with which it has been projected.

Now suppose a projectile is projected with an initial velocity $u$ at an angle $\theta$ w.r.t the horizontal,so we can discuss the effect of gravity after breaking the velocity into two perpendicular components.

As gravity will affect the vertical component only.

So,vertical component of its velocity is $u \sin \theta$,so the projectile will keep on moving up,untill it's upward velocity becomes zero due to the downward direction of gravitational force acting on it.

So, at the highest point of its motion,the projectile has no vertical component of velocity,only horizontal component of velocity exists.

After that the projectile starts coming down being accelerated by gravity.So it's height decreases and at a time it reaches the ground.

So where,the horizontal component of velocity pushes it forwards,vertical component of velocity pushes it upwards,but it comes back to the ground just because of the gravitational force.

• 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).

• 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.