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).
Combine the horizontal and vertical equations of motion for a projectile moving against a
linear drag force, see a previous task, to (A) obtain an equation of the trajectory of the
projectile, i.e., z(x). (B) Obtain an equation for the RANGE (i.e., maximum horizontal
distance reached) of the projectile. (C) Compare the range equation with an equation for
range obtained in the case of vanishing drag force. Discuss the differences.
To solve the given problem, we need to apply the equations of motion and incorporate the drag force into the dynamics. Let's break down each part of the problem.
(A) Vertical Motion:
We have a vertical drop with only the z-component of velocity, vz. The equation of motion in the vertical direction, taking into account the drag force, is given by:
m * dvz/dt = -mg - b * vz,
where m is the mass of the projectile, g is the acceleration due to gravity, and b is the drag coefficient.
(a) Solving the above differential equation, we obtain the velocity as a function of time:
vz(t) = (g/b) * (1 - exp(-b*t/m)),
where vz(0) = 0.
(b) To find the vertical position change, we integrate the velocity equation with respect to time:
z(t) = (m/g) * ((g/b)*t + (m/b^2) * (exp(-b*t/m) - 1)) + H,
where z(0) = H is the initial height.
(B) Horizontal Motion:
We now consider only the horizontal motion with the drag force. The equation of motion in the horizontal direction is given by:
m * dvx/dt = -b * vx,
where vx is the horizontal component of velocity.
(a) Solving the above differential equation, we get the velocity as a function of time:
vx(t) = vox * exp(-b*t/m),
where vx(0) = vox is the initial horizontal velocity.
(b) The horizontal position change can be obtained by integrating the velocity equation with respect to time:
x(t) = (m/b) * (vox - vx(t)),
where x(0) = 0.
Combining the horizontal and vertical equations of motion:
(A) To obtain the equation of the projectile trajectory, z(x), we can substitute the horizontal position, x(t), from the horizontal motion equation into the vertical position, z(t), from the vertical motion equation.
(B) To find the range, i.e., the maximum horizontal distance reached by the projectile, we need to determine the value of x when z(x) equals zero.
(C) To compare the range equation with the case of vanishing drag force, we simply need to analyze the range equation obtained in the absence of the drag force.
By following the steps outlined above, we can derive the equations and discuss the differences in the range in cases with and without drag force.