We consider a projectile motion against a linear drag force D = −b∗v, where v is the velocity
of the projectile.
(A) Suppose only a vertical drop (in z-direction), v = vz, from an initial height H with
an initial velocity voz = 0. Obtain the corresponding equations for (a) velocity vz(t), (b)
vertical position change of the projectile z(t).
(B) Consider now only a horizontal motion (with drag) v = vx, from an initial height H and
with an initial horizontal velocity vox. Obtain the corresponding equations for (a) velocity
vx(t), (b) horizontal position change of the projectile x(t).
To obtain the equations for the vertical motion (part A) and horizontal motion (part B) of the projectile with linear drag force, we need to consider the equations of motion under the influence of the drag force.
(A) Vertical Motion (v = vz):
The equation of motion in the vertical direction is given by Newton's second law: F = ma, where F is the net force, m is the mass of the projectile, and a is the acceleration. In this case, the net force is the gravitational force mg acting downward and the drag force D = -bvz acting upward. We can write this as:
mg + D = ma
Since we are considering only the vertical motion, we can focus on the z-direction:
mg - bvz = m(dvz/dt)
Rearranging the equation, we get:
mg - bvz = m(dvz/dt)
dvz/dt = -g - (b/m)vz
This is a first-order linear ordinary differential equation. To solve it, we need to find the velocity vz(t) and vertical position change z(t) as functions of time.
To obtain the equation for velocity vz(t), we can separate variables and integrate both sides:
dvz/(g + (b/m)vz) = -dt
Integrating both sides with respect to vz, we get:
(m/b)ln(g + (b/m)vz) = -t + C
(ln(g + (b/m)vz))/(g/b + vz) = -(1/m)t + C
Here, C is the constant of integration. We can solve for C using the initial condition that voz = 0 at t = 0:
(ln(g + (b/m)vz))/(g/b + vz) = 0 + C
(ln(g + (b/m)vz))/(g/b + vz) = C
To solve for vz(t), we can exponentiate both sides:
g + (b/m)vz = Ce^(-t/mg)
vz(t) = (Ce^(-t/mg) - g)(m/b)
Now, to find the vertical position change z(t), we can integrate the velocity vz(t) with respect to time:
z(t) = ∫vz(t)dt
z(t) = ∫[(Ce^(-t/mg) - g)(m/b)]dt
z(t) = [(Cm/b)e^(-t/mg) - gt](m/b)
(B) Horizontal Motion (v = vx):
Similar to the vertical motion, the horizontal motion of the projectile is governed by the equation of motion:
ma = D
ma = -bv
Since we are considering only the horizontal motion, we can focus on the x-direction:
- bvx = m(dvx/dt)
Rearranging the equation, we get:
dvx/dt = -(b/m)vx
Again, this is a first-order linear ordinary differential equation. To solve it, we need to find the velocity vx(t) and horizontal position change x(t) as functions of time.
The equation for velocity vx(t) can be obtained by separating variables and integrating both sides:
dvx/vx = -(b/m)dt
Integrating both sides with respect to vx, we get:
ln(vx) = -(b/m)t + C
vx(t) = Ce^(-t(b/m))
To find the horizontal position change x(t), we can integrate the velocity vx(t) with respect to time:
x(t) = ∫vx(t)dt
x(t) = ∫(Ce^(-t(b/m)))dt
x(t) = (-Cm/b)e^(-t(b/m))
So, the equations for (a) velocity vz(t) and (b) vertical position change z(t) are:
(a) vz(t) = (Ce^(-t/mg) - g)(m/b)
(b) z(t) = [(Cm/b)e^(-t/mg) - gt](m/b)
The equations for (a) velocity vx(t) and (b) horizontal position change x(t) are:
(a) vx(t) = Ce^(-t(b/m))
(b) x(t) = (-Cm/b)e^(-t(b/m))