Vipul: Created page with "{{mechanics scenario}} We consider here a situation where a particle is given an initial speed into the air in some direction and is then left to itself with no forces acting on..."

2011-08-24T11:57:49Z

Created page with "{{mechanics scenario}} We consider here a situation where a particle is given an initial speed into the air in some direction and is then left to itself with no forces acting on..."

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{{mechanics scenario}}

We consider here a situation where a particle is given an initial speed into the air in some direction and is then left to itself with no forces acting on it other than [[acceleration due to gravity]], until it comes into contact with a solid surface (such as the ground). We ignore [[air drag]].

The particle being released is termed a ''projectile'' and the motion it exhibits is termed ''projectile motion''.

==When the projectile is fired from a horizontal surface and there is no other surface in sight==

We let <math>u</math> be the initial speed and <math>\alpha</math> be the angle with the horizontal (in the upward direction) at which the projectile is fired.

We take coordinates with origin at the point the projectile is fired from, with the <math>x</math>-direction the horizontal direction of firing (positive along the direction of firing), and the <math>y</math>-direction the vertical direction (upward positive) and do the analysis.

===Dynamics===

The component of acceleration along the horizontal direction is zero, i.e., we get:

<math>\frac{d^2x}{dt^2} = 0</math>

The component of acceleration along the vertical direction is <math>g</math> downward, so we get:

<math>\frac{d^2y}{dt^2} = -g</math>

===Kinetics===

We use the dynamics results and the initial conditions:

<math>\frac{dx}{dt}|_{t = 0} = u \cos \alpha, \qquad x |_{t = 0} = 0</math>

to obtain:

<math>x(t) = ut \cos alpha</math>

Similarly, we get:

<math>y(t) = ut \sin \alpha - \frac{1}{2}gt^2</math>

Also, the horizontal and vertical components of velocity are <math>x'(t) = u\cos \alpha</math> and <math>y'(t) = u \sin \alpha - gt</math> respectively.
===Key quantities===

We have the following formulas for key quantities:

{| class="sortable" border="1"
! Quantity !! Value !! Explanation
|-
| Time taken to attain maximum height || <math>\frac{u\sin \alpha}{g}</math> || In the expression <math>y'(t) = u\sin \alpha - gt</math>, set <math>y'(t) = 0</math>.
|-
| Value of maximum height || <math>\frac{u^2\sin^2\alpha}{2g}</math> || Can be seen by plugging in time from above into the expression for <math>y(t)</math>. Can also be seen using a conservation of energy argument.
|-
| Time taken to hit the ground again || <math>\frac{2u\sin \alpha}{g}</math> || In the expression <math>y(t) = ut \sin \alpha - \frac{1}{2}gt^2</math>, set <math>y(t) = 0</math>.
|-
| Horizontal distance traveled before hitting the ground again || <math>\frac{u^2\sin(2\alpha)}{g}</math> || Plug in time value above into expression for <math>x(t)</math>, use identity <math>\sin(2\alpha) = 2\sin \alpha \cos \alpha</math>
|}

===Choosing the angle appropriately===

We note from the above that:

* For a given initial speed <math>u</math>, the maximum vertical height, as well as maximum ''time'' taken to attain the maximum vertical height, occurs when <math>\alpha = \pi/2</math>, i.e., the projectile is fired vertically.
* For a given initial speed <math>u</math>, the maximum horizontal distance traversed occurs when <math>\alpha = \pi/4</math>.

Projectile motion - Revision history

Vipul: Created page with "{{mechanics scenario}} We consider here a situation where a particle is given an initial speed into the air in some direction and is then left to itself with no forces acting on..."