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Πλαγία Βολή

- Ένα είδος βολής.

## ΕτυμολογίαEdit

Η ονομασία "Βολή" σχετίζεται ετυμολογικά με την λέξη "[[]]".

## ΕισαγωγήEdit

Projectile motion is a form of motion in which an object or particle (in either case referred to as a projectile) is thrown near the Earth's surface, and it moves along a curved path under the action of gravity only. The implication here is that air resistance is negligible, or in any case is being neglected in all of these equations.

The only force of significance that acts on the object is gravity, which acts downward to cause a downward acceleration. Because of the object's inertia, no external horizontal force is needed to maintain the horizontal velocity of the object.

## ΑνάλυσηEdit

### The initial velocity Edit

Let the projectile be launched with an initial velocity $\mathbf{v} (0) \equiv \mathbf{v}_0$, which can be expressed as the sum of horizontal and vertical components as follows:

$\mathbf{v}_0 = v_{0x}\mathbf{i} + v_{0y}\mathbf{j}$.

The components $v_{0x}$ and $v_{0y}$ can be found if the initial launch angle, $\theta$, is known:

$v_{0x} = v_0\cos\theta$,
$v_{0y} = v_0\sin\theta$.

### Kinematic quantities of projectile motion Edit

In projectile motion, the horizontal motion and the vertical motion are independent of each other; that is, neither motion affects the other. This is the principle of compound motion established by Galileo in 1638.[1]

### Acceleration Edit

Since there is only acceleration in the vertical direction, the velocity in the horizontal direction is constant, being equal to $\mathbf{v}_0 \cos\theta$. The vertical motion of the projectile is the motion of a particle during its free fall. Here the acceleration is constant, being equal to $g$.[2] The components of the acceleration are:

$a_x = 0$,
$a_y =- g$.

### Velocity Edit

The horizontal component of the velocity of the object remains unchanged throughout the motion. The downward vertical component of the velocity increases linearly, because the acceleration due to gravity is constant. The accelerations in the $x$ and $y$ directions can be integrated to solve for the components of velocity at any time $t$, as follows:

$v_x=v_0 \cos(\theta)$,
$v_y=v_0 \sin(\theta) - gt$.

The magnitude of the velocity (under the Pythagorean theorem, also known as the triangle law):

$v=\sqrt{v_x^2 + v_y^2 \ }$.

### Displacement Edit

At any time $t$, the projectile's horizontal and vertical displacement are:

$x = v_0 t \cos(\theta)$,
$y = v_0 t \sin(\theta) - \frac{1}{2}gt^2$.

The magnitude of the displacement is:

$\Delta r=\sqrt{x^2 + y^2 \ }$.

Consider the equations,

$x = v_0 t \cos(\theta) , y = v_0 t\sin(\theta) - \frac{1}{2}gt^2$.

If t is eliminated between these two equations the following equation is obtained:

$y=\tan(\theta) \cdot x-\frac{g}{2v^2_{0}\cos^2 \theta} \cdot x^2$.

Since $g$, $\theta$, and $\mathbf{v}_0$ are constants, the above equation is of the form

$y=ax+bx^2$,

in which $a$ and $b$ are constants. This is the equation of a parabola, so the path is parabolic. The axis of the parabola is vertical.

If the projectile's position (x,y) and launch angle (θ or α) are known, the initial velocity can be found solving for $v_{0}$ in the aforementioned parabolic equation:

$v_0 = \sqrt{{x^2 g} \over {x \sin 2\theta - 2y \cos^2\theta}}$.

### Time of flight or total time of the whole journey Edit

The total time $t$ for which the projectile remains in the air is called the time of flight.

$y = v_0 t \sin(\theta) - \frac{1}{2}gt^2$

After the flight, the projectile returns to the horizontal axis (x-axis), so y=0

$0 = v_0 t \sin(\theta) - \frac{1}{2}gt^2$
$v_0 t \sin(\theta) = \frac{1}{2}gt^2$
$v_0 \sin(\theta) = \frac{1}{2}gt$
$t = \frac{2 v_0 \sin(\theta)}{g}$

Note that we have neglected air resistance on the projectile.

## Maximum height of projectile Edit

The greatest height that the object will reach is known as the peak of the object's motion. The increase in height will last until $v_y=0$, that is,

$0=v_0 \sin(\theta) - gt_h$.

Time to reach the maximum height(h):

$t_h = \frac{v_0 \sin(\theta)}{g}$.

From the vertical displacement of the maximum height of projectile:

$h = v_0 t_h \sin(\theta) - \frac{1}{2} gt^2_h$
$h = \frac{v_0^2 \sin^2(\theta)}{2g}$ .

### Relation between horizontal range and maximum height Edit

The relation between the range $R$ on the horizontal plane and the maximum height $h$ reached at $\frac{t_d}{2}$ is:

$h = \frac{R\tan\theta}{4}$

Απόδειξη:

$h = \frac{v_0^2\sin^2\theta}{2g}$
$R = \frac{v_0^2\sin2\theta}{g}$
$\frac{h}{R} = \frac{v_0^2\sin^2\theta}{2g}$ × $\frac{g}{v_0^2\sin2\theta}$
$\frac{h}{R} = \frac{\sin^2\theta}{4\sin\theta\cos\theta}$

$h = \frac{R\tan\theta}{4}$.

## Maximum distance of projectile Edit

It is important to note that the range and the maximum height of the projectile does not depend upon its mass. Hence range and maximum height are equal for all bodies that are thrown with the same velocity and direction. The horizontal range d of the projectile is the horizontal distance it has traveled when it returns to its initial height (y = 0).

$0 = v_0 t_d \sin(\theta) - \frac{1}{2}gt_d^2$.

Time to reach ground:

$t_d = \frac{2v_0 \sin(\theta)}{g}$.

From the horizontal displacement the maximum distance of projectile:

$d = v_0 t_d \cos(\theta)$,

so[3]

$d = \frac{v_0^2}{g}\sin(2\theta)$.

Note that $d$ has its maximum value when

$\sin 2\theta=1$,

which necessarily corresponds to

$2\theta=90^\circ$,

or

$\theta=45^\circ$.

### Application of the work energy theorem Edit

According to the work-energy theorem the vertical component of velocity is:

$v_y^2 = (v_0 \sin \theta)^2-2gy$.

## ΥποσημειώσειςEdit

1. Galileo Galilei, Two New Sciences, Leiden, 1638, p.249
2. The $g$ is the acceleration due to gravity. ($9.81 m/s^2$ near the surface of the Earth).
3. $2\cdot\sin(\alpha)\cdot\cos(\alpha) = \sin(2\alpha)$

## ΙστογραφίαEdit

Κίνδυνοι Χρήσης

Αν και θα βρείτε εξακριβωμένες πληροφορίες
σε αυτήν την εγκυκλοπαίδεια
ωστόσο, παρακαλούμε να λάβετε σοβαρά υπ' όψη ότι
η "Sciencepedia" δεν μπορεί να εγγυηθεί, από καμιά άποψη,
την εγκυρότητα των πληροφοριών που περιλαμβάνει.

"Οι πληροφορίες αυτές μπορεί πρόσφατα
να έχουν αλλοιωθεί, βανδαλισθεί ή μεταβληθεί από κάποιο άτομο,
η άποψη του οποίου δεν συνάδει με το "επίπεδο γνώσης"
του ιδιαίτερου γνωστικού τομέα που σας ενδιαφέρει."

Πρέπει να λάβετε υπ' όψη ότι
όλα τα άρθρα μπορεί να είναι ακριβή, γενικώς,
και για μακρά χρονική περίοδο,
αλλά να υποστούν κάποιο βανδαλισμό ή ακατάλληλη επεξεργασία,
ελάχιστο χρονικό διάστημα, πριν τα δείτε.

Επίσης,
(όχι μόνον, της Sciencepedia
αλλά και κάθε διαδικτυακού ιστότοπου (ή αλλιώς site)),
αν και άκρως απαραίτητοι,
είναι αδύνατον να ελεγχθούν
(λόγω της ρευστής φύσης του Web),
και επομένως είναι ενδεχόμενο να οδηγήσουν
σε παραπλανητικό, κακόβουλο ή άσεμνο περιεχόμενο.
Ο αναγνώστης πρέπει να είναι
εξαιρετικά προσεκτικός όταν τους χρησιμοποιεί.

- Μην κάνετε χρήση του περιεχομένου της παρούσας εγκυκλοπαίδειας
αν διαφωνείτε με όσα αναγράφονται σε αυτήν

>>Διαμαρτυρία προς την wikia<<

- Όχι, στις διαφημίσεις που περιέχουν απαράδεκτο περιεχόμενο (άσεμνες εικόνες, ροζ αγγελίες κλπ.)