a gun is fired with muzzle velocity v0 = 1000 m/s straight up along the z-axis. While moving through the air the Earth's gravity pulls on the bullet, which has mass m = 20 grams and the bullet's air resistance force can be approximated by the relationship Fresist = -cv^2 where c is a constant = 2*10^-5 kg/m and where v=v(t) is the bullet's instantaneous speed in flight. Determine the expressions for the bullet's speed as a function of height z above the firing position during both the up-leg of its travel and during the down leg
define positive up
Vi = + 1000
going up
F = - m g - c v^2 = m dv/dt
dv/dt = -g -(c/m) v^2
v = Vi - g t -(c/m)integral of v^2 dt
z = Vi t - (1/2) g t^2 - (c/m) double integral of v^2 dt
I do not think there is an analytical solution. You must resort to numerical integration.
To determine the expressions for the bullet's speed as a function of height during both the up-leg and down-leg of its travel, we need to consider the forces acting on the bullet.
During the up-leg (which is when the bullet is moving upward from the firing position), the forces acting on the bullet are gravity and air resistance. The gravity force is given by Fgravity = -mg, where m is the mass of the bullet and g is the acceleration due to gravity (approximately 9.8 m/s^2). The air resistance force is given by Fresist = -cv^2, where c is a constant.
Using Newton's second law of motion (F = ma), we can write the equation of motion for the bullet during the up-leg:
Fgravity + Fresist = ma
Since the bullet is moving along the z-axis, the equation can be written as:
-mg - cv^2 = m(dv/dt)
Rearranging this equation, we get:
- dv/v^2 = (g + (c/m)v) dt
Now, we can separate the variables and integrate both sides:
∫- dv/v^2 = ∫(g + (c/m)v) dt
Integrating, we get:
(1/v) = (g + (c/m)v)t + A
Where A is a constant of integration. At t = 0 (when the bullet is fired), the velocity v = v0 (initial velocity of 1000 m/s). So, we can substitute these values into the equation to solve for A:
(1/v0) = (g + (c/m)v0)(0) + A
Simplifying, we find that A = (1/v0).
Substituting this value of A back into the integrated equation, we get:
(1/v) = (g + (c/m)v)t + (1/v0)
Now, we can solve for v as a function of time during the up-leg of the bullet's travel:
v = 1 / [(g + (c/m)v0)t + (1/v0)]
To determine the expression for the bullet's speed as a function of height z, we can integrate the velocity equation with respect to time. The speed of the bullet is the absolute value of velocity, so the final expression would be:
speed during up-leg = |v| = 1 / |(g + (c/m)v0)t + (1/v0)|
During the down-leg (when the bullet is moving downward), the forces acting on the bullet remain the same (gravity and air resistance), but the direction of these forces changes. However, we can use the same equation as before, as the gravitational force is still in the opposite direction of the bullet's motion. The only change will be the limits of integration when solving for A (since we are considering the down-leg of travel).
Therefore, the expression for the bullet's speed as a function of height z during the down-leg will be the same as the expression for the up-leg:
speed during down-leg = |v| = 1 / |(g + (c/m)v0)t + (1/v0)|