Ballistics Trajectory- Physics Equations

[Suicidesoldier#1]

Ballistics Trajectory- Physics Equations

While I understand the basic concepts of Trajectories of object, such that vertical acceleration, such as gravity applying force in the negative direction and horizontal forces applying force in the positive direction, and how Sin and Cos help determine height and distance for a parabolic arc, I'm still struggling with the mathematics as a whole.

I understand the basic concepts, but what I'm having trouble with is understanding the mathematics. I have, so far, yet to find a website or a textbook that explains how to preform advanced ballistic trajectories, disregarding air resistance or not, mathematically.

I understand that Gravity is explained with 1/2G^2 and how sin and cos are taken by the radian, and then multiplied into the equation, but what I'm having trouble with is determining how you find out any of the information at all with a singular equation.

Can anybody help me with this?  

[zz1000zz]

I assume you start off with a value for velocity and angle of inclination. I also assume there is no air resistance.

Supposing you those two values, figuring out how far a projectile will travel is fairly simple. First, break the velocity into its components, the horizontal velocity and the vertical velocity.

vx = v * cos(θ)
vy = v * sin(θ)

You should know the formula
vf = vi + a * t.
You should also know the time it takes an object to reach the highest point of its curve is the same time it takes for it to fall back down. Finally, you should know the velocity at the top of the curve is zero. Combine these things, apply a little algebra, and you get:

t = 2 * -vy/a

Plug in your value for gravity (remember, it should be negative) and the vertical velocity, solve and you have your value for time. Multiply that by the horizontal velocity, and you have the distance the projectile will travel.

As long as you don't have to worry about air resistance, ballistics trajectory shouldn't be hard. Once you break the horizontal and vertical components apart, it is basically just your normal linear physics problems.  

[Suicidesoldier#1]

zz1000zz

I assume you start off with a value for velocity and angle of inclination. I also assume there is no air resistance.

Supposing you those two values, figuring out how far a projectile will travel is fairly simple. First, break the velocity into its components, the horizontal velocity and the vertical velocity.

vx = v * cos(θ)
vy = v * sin(θ)

You should know the formula
vf = vi + a * t.
You should also know the time it takes an object to reach the highest point of its curve is the same time it takes for it to fall back down. Finally, you should know the velocity at the top of the curve is zero. Combine these things, apply a little algebra, and you get:

t = 2 * -vy/a

Plug in your value for gravity (remember, it should be negative) and the vertical velocity, solve and you have your value for time. Multiply that by the horizontal velocity, and you have the distance the projectile will travel.

As long as you don't have to worry about air resistance, ballistics trajectory shouldn't be hard. Once you break the horizontal and vertical components apart, it is basically just your normal linear physics problems.

*blink*

REALLY? eek

Oh my gosh thank you! I've never been able to put all those concepts together... sweatdrop

So horizontal distance is determined by the original horizontal velocity? Woot!

Do you have anything about air resistance; or is that uber complicated and not just a simple multiplication thing.  

[zz1000zz]

The problem with air resistance is it is not a constant. It varies dependent upon speed (as well as other factors), so it winds up being difficult to calculate.

With that said, you may be able to approximate air resistance in some instances.  

[VorpalNeko]

The horizontal velocity doesn't change because the acceleration is purely vertical (i.e., gravity is only vertical).

Air resistance is comparatively very complicated. I don't think I can explain it simply enough if you've trouble with the frictionless case, but if you're curious, I can just give you the solution.

If one assumes that it is proportional to velocity (which is valid in some regimes, though hardly all), x" = -kx' (i.e., acceleration is proportional to velocity) has the solution
[1] x(t) = x_0 + (v_x/k)[ 1 - exp(-kt) ]
where x_0 and v_x are the initial position and initial x-velocity, respectively. In the vertical direction, y" = -(g+ky') (i.e., acceleration is a constant gravity plus the same velocity-proportional term) has the solution
[2] y(t) = y_0 - gt/k + [(g + k v_y)/k²][ 1 - exp(-kt) ]
Which looks singular as k→0, but really isn't, for if the exponential is expanded into a Maclaurin series, all negative-power k terms cancel. An interesting observation that can be made here: if t>>1, then the -gt/k term dominates over the rest, so g/k is the terminal velocity of the object.

The case where the resistance follows the drag equation (proportional to velocity-squared rather than velocity) also has an analytic solution, but it's badly behaved and invalid for many initial conditions (the difficulty is that a -ky'² term can fail to correctly respond to the direction of the velocity, and the more correct -ky'³/|y'| does not have an analytic solution).  

[bigmanlittlepackage]

I don't post here much, I'm not half as magnificently minded as anybody here, so writing this I'll probably be complete on the wrong track from what the OP is getting at. But here goes.

Where I'm from. We use SUVAT. S(displacement)U(initial vel.)V(velocity)A(acceleration)T(time).

S=UT + 1/2AT^2
V=U+AT
V^2=U^2+2AS
A=(V-U)/T
V=S/T

Works for me every time. In exam papers they state which value of gravity to use. When finding the components of vector triangles using trig.

However judging from some of the answers here I'm just a minnow in a stream compared to you sharks.