Analyze the flight of a projectile moving under the influence of gravity (mg) and linear, i.e., laminar-flow drag (fD = -bv), with b constant). Neglect (phase I); assume the projectile is launched from mthe origin at speed v1 and angle theta1 above the horizontal. WARNING: Neither the results for the drag-free projectile nor those for the projectile with turbulent-flow drag can be applied directly to this problem; you must go back to basic principles.
a. Determine the velocity components vx and vy and the coordinates x and y, of the projectile as functions of time.
b. Ccalculate the horizontal position of the peak of the projectile’s trayectory (xpeak) and the height for the peak (ypeak). Remember the WARNING!
To analyze the flight of a projectile moving under the influence of gravity and linear drag, we can break down the problem into two components: the horizontal (x) and vertical (y) motion. Let's begin with determining the velocity components vx and vy and the coordinates x and y as functions of time.
a. Horizontal Motion (x-component):
In the absence of any external forces, the horizontal velocity (vx) remains constant throughout the motion. Hence, vx = v1 * cos(theta1).
The horizontal displacement (x) of the projectile can be calculated using the equation:
x = v1 * cos(theta1) * t
b. Vertical Motion (y-component):
Considering the influence of gravity (mg) and the linear drag (fD = -bv), the equation of motion for the vertical component is given by:
m * dv/dt = -mg - bv
Simplifying the equation, we can write it as:
dv/dt = -(g + (b/m) * v)
To solve this differential equation, we can use the separation of variables method:
dv / (g + (b/m) * v) = -dt
Integrating both sides, we get:
(m/b) * ln(abs(g + (b/m)*v)) = -t + C
Applying initial conditions, we can find the value of the constant C.
When t = 0, v = v1 * sin(theta1).
Hence,
(m/b) * ln(abs(g + (b/m) * (v1 * sin(theta1)))) = C
Now, rearranging the equation, we have:
ln(abs(g + (b/m) * v)) = - (b/m) * t + ln(abs(g + (b/m) * v1 * sin(theta1)))
Taking the exponential of both sides, we get:
abs(g + (b/m) * v) = (g + (b/m) * v1 * sin(theta1)) * exp(-(b/m) * t)
Since we've assumed the drag force is linear and the drag coefficient is constant, the velocity will approach terminal velocity (v_terminal) over time, where the drag force and gravitational force balance each other. We can express terminal velocity as:
v_terminal = g * m / b
Therefore, we can rewrite the equation as:
g + (b/m) * v = (g + (b/m) * v1 * sin(theta1)) * exp(-(b/m) * t)
Rearranging the equation, we get:
v = v_terminal + (v1 * sin(theta1) - v_terminal) * exp(-(b/m) * t)
Now, integrating the equation for velocity with respect to time, we can find the equation for vertical displacement (y):
y = y0 + (v_terminal * t) - ((v_terminal - v1 * sin(theta1)) * (m/b)) * (1 - exp(-(b/m) * t))
This equation provides the vertical position of the projectile as a function of time.
b. Determining the peak's horizontal position (xpeak) and height (ypeak):
To find the horizontal position of the peak (xpeak), we can set the vertical velocity (vy) equal to zero, since the projectile reaches its maximum height at that point. Solving for t in the equation vy = vy0 - g * t = 0, we find:
t = vy0 / g
Substituting this value of t into the equation for horizontal motion, we can calculate xpeak. Hence:
xpeak = v1 * cos(theta1) * (vy0 / g)
To determine the height of the peak (ypeak), we substitute the value of t = vy0 / g into the equation for vertical displacement. Thus:
ypeak = y0 + (v_terminal * (vy0 / g)) - ((v_terminal - v1 * sin(theta1)) * (m/b)) * (1 - exp(-(b/m) * (vy0 / g)))
Remember the warning mentioned earlier - the results for the drag-free or turbulent-flow drag cannot be applied directly here as the linear drag equations must be derived from basic principles.
By following the steps outlined above, you can calculate the velocity components, coordinates, peak horizontal position, and peak height of a projectile moving under the influence of gravity and linear drag.
To analyze the flight of the projectile, we will break it down into different components and consider the forces acting on it.
a. Velocity Components and Projectile Coordinates:
The projectile experiences two forces: gravity (mg) and drag force (fD = -bv).
Let's consider the x-axis (horizontal direction) and the y-axis (vertical direction):
x-axis:
There is no force acting on the x-axis, so the projectile's velocity in the x-direction (vx) remains constant. The initial velocity in the x-direction (v1x) can be found using trigonometry:
vx = v1 * cos(theta1)
The projectile's position in the x-direction (x) can be determined using the equation of motion:
x = v1 * cos(theta1) * t
y-axis:
The projectile is affected by gravity and drag force in the y-direction. The y-component of initial velocity (v1y) can also be found using trigonometry:
v1y = v1 * sin(theta1)
Let's consider the vertical motion separately:
Summing up the forces in the y-direction:
- Drag force (fD) = -bv
- Gravity (mg) = -mg
Using Newton's second law, we can write the equation of motion in the y-direction:
m * dv/dt = -bv - mg
Rearranging and separating variables, we get:
m * dv/(bv + mg) = -dt
Integrating both sides of the equation, we obtain:
m * ln|bv + mg| = -t + C1
Simplifying the equation and solving for v, we get:
v = (-mg/b) + C2 * exp(-bt/m)
Applying initial conditions, when t = 0, v = v1y:
v1y = (-mg/b) + C2 * exp(-b*0/m)
v1y = (-mg/b) + C2
Solving for C2:
C2 = v1y + (mg/b)
Substituting the value of C2 back into the equation, we get:
v = (-mg/b) + (v1y + (mg/b)) * exp(-bt/m)
Integrating both sides once more, we get:
y = (-mg/b)t + (v1y + (mg/b))(m/b)(1 - exp(-bt/m))
Now, we can substitute the expressions for v1y and b into the equation:
y = (-mg/b)t + [(v1 * sin(theta1)) + (mg/b)][m/b][1 - exp(-bt/m)]
b. Calculation of xpeak and ypeak:
To find the horizontal position of the peak of the trajectory (xpeak), we set vy = 0:
0 = (-mg/b)t + [(v1 * sin(theta1)) + (mg/b)][m/b][1 - exp(-bt/m)]
Simplifying the equation, we get:
t = (m/b) * ln[(v1 * sin(theta1)) * (b/m) / (mg + b * (v1 * sin(theta1)))]
Substituting this value of t back into the equation for x, we can find xpeak.
To calculate the height of the peak (ypeak), we substitute t = xpeak/v1x into the equation for y.
Remember, these calculations are based on the given assumptions and should not be directly applied to other scenarios.