Ballistics

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The study of projectiles and their motion.

1 How Ballistics Works
2 Projectile Fired Vertically
- 2.1 Freefall
- 2.2 Firing Directly Upwards
3 Projectile Fired Horizontally
4 Projectile Fired at an Angle
- 4.1 Projectile Fired from Level Surface

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How Ballistics Works

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Projectile Fired Vertically

A projectile fired vertically with negligable air resistance and in uniform gravity obeys the law

$h = vt - \frac{gt^2}{2}$

Where $h$ is the height relative to the starting position, $v$ the initial upward velocity of the projectile, $t$ time and $g$ the acceleration due to gravity, generally taken as 9.8m/s/s on Earth.

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Freefall

An important special case of this is a projectile falling freely, in this case $v = 0$ and $h = - \frac{gt^2}{2}$ . Rearranging this, such a projectile takes $t = \sqrt{\frac{2h}{g}}$ to fall a distance $h$ .

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Firing Directly Upwards

A projectile fired directly upward at $v$ reaches a height of $v 2 / 2 g$ . Conversely, a projectile that reaches a height $h$ had an upward velocity of $\sqrt{2gh}$ . Such a projectile also takes a time $v / g$ to reach its maximum height.

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Projectile Fired Horizontally

The key observation in understanding this case is that motion can be split into horizontal and vertical components (or, in fact, components in any two given directions) that are completely independent. Eg, something with horizontal velocity 10m/s will travel 10m horizontally in 1s regardless of what its vertical velocity or position is.

So a body fired horizontally (such as off a table) will take the same amount of time to reach the ground however fast it travels horizontally (assuming air resistance is negligable). The horizontal velocity is approximately constant, so the horizontal distance covered is just the time taken to reach the ground mulitplied by the horizontal velocity.

Taking a formula from the above section, the horizontal distance covered by the projectile is

$s_h = v_h \sqrt{\frac{2h}{g}}$

Rearranging, if the horizontal distance is measured, then the velocity of the projectile was

$v_h = s_h \sqrt{\frac{g}{2h}}$

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Projectile Fired at an Angle

A projectile fired at an angle $θ$ from the horizontal at a velocity $v$ has vertical velocity $v v = v sinθ$ and horizontal velocity $v h = v cosθ$ .

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Projectile Fired from Level Surface

This projectile spends a time $t = 2 v v / g$ in the air and so covers a distance $s = 2 v_v \times v_h / g$ before it lands, or (with a little trigonometry) $s = \frac {v^2 sin2\theta}{g}$ .