so I got a question about projectile motion
take the example of shooting of a plastic air rocket into the air and asked to predict its landing point and how high it goes...
ok I know how to do this...
the thing is now were not neglecting air resistence and I don't know what to do...
I know how to prove the derivation of this equation for distance traveled in the x direction
X = Xo + Vo t + 2^-1 a t^2
ok now sense we were once negelcting air resistance I was always told to use this equation because the velocity stays the same and therefore there is no acceleration
X = Xo + Vo t
because the acceleration is zero taking out that whole set of terms
ok now lets not neglect air resistance...
I know I would have to use every component of this equation except for the Xo which we can let be zero
X = Xo + Vo t + 2^-1 a t^2
take out zero term
x = Vo t + 2^-1 a t^2
ok now what???
I obviously have to do something else...
How do I find the acceleration???
We are given the launch angle but have to derive the launch velocity :(
how does the drag force effect motion in the y direction???
I got to use the same equation I know but don't know how to find the acceleartion
also what's the range equation it's something like
x = f (v, /0)
were the "/0" is like a o with a slash mark thingy with it... what does this symbol represent and how do I derive this equation (I just need the start) and what does the f represent
also I was always told to use this equation when the velocity is small through the medium
Fd = -bv
were this is just a proportion
also that other formula that is suppose to be used when there is a high velocity
Fd = 2^-1 p v^2 A Cd v
Ok what is exactly meant by high and low velocity and how do you derive the equation for a high velocity all I've been told was the equation never taught how it was derived and don't know were to start
I just would like to make sure of something
an object experiances a drag force when it travels through a medium like air or water...
so dosen't every thing experience a drag force that is in motion because its moving through air??
So like if a person was walking do they experiance a drag force?
I would think so... just making sure
There is always a drag force when a body moves through a fluid, except in some rare exotic situations involving superfluids like liquid helium. To treat the drag force, you need to know how it varies with velocity. At very slow speeds the resistance is proportional to velocity, but at "normal" speeds it varies with the square of velocity. The drag force needs to be included in the differential equation of motion. (calculus is needed)
To account for air resistance in projectile motion, you need to consider the components of acceleration due to air resistance separately in the x and y directions. Let's break it down step by step:
1. Finding the acceleration in the x-direction:
- Air resistance only acts horizontally, so it does not affect the vertical motion.
- The force of air resistance is proportional to the velocity in the x-direction, usually modeled as F_air = -k * v_x.
- From Newton's second law (F = ma), we can equate this force to the mass of the rocket times the acceleration in the x-direction (a_x): -k * v_x = m * a_x.
- Dividing both sides by the mass gives you the equation for acceleration in the x-direction: a_x = -k * v_x / m.
2. Determining the effect of drag force on motion in the y-direction:
- In the y-direction, air resistance acts vertically upward opposite to the gravitational force.
- The force of air resistance in this case is proportional to the velocity in the y-direction, modeled as F_air = -k * v_y.
- Again using Newton's second law, the sum of forces in the y-direction is given by F_net = F_air + mg (where mg is the weight of the rocket).
- The acceleration in the y-direction is then determined by dividing the net force by the mass: a_y = (F_air + mg) / m.
3. Range equation:
The equation you mentioned, x = f(v, ø), represents the range equation for projectile motion when air resistance is considered. In this equation:
- x is the horizontal distance traveled by the projectile (range).
- v is the launch velocity.
- ø is the launch angle.
- The symbol "/0" represents the ratio of the vertical component of the launch velocity (v_y) to the horizontal component of the launch velocity (v_x) and is denoted as v_y / v_x. This ratio helps derive the expression for the range equation.
Deriving the full range equation is quite involved, but the basic idea is to solve the differential equations for x(t) and y(t) using the initial conditions (initial velocity, launch angle, and position). The final expression will involve trigonometric functions and will take into account the acceleration due to air resistance.
It's essential to note that in practical scenarios, the exact solutions are often quite complex. In some cases, numerical methods or computer simulations may be needed to accurately predict projectile motion with air resistance.