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!
a) To analyze the flight of the projectile, we can break down the initial velocity v1 into its horizontal (vx) and vertical (vy) components. The horizontal component remains constant throughout the motion, while the vertical component changes under the influence of gravity and drag.
1. Horizontal Component:
The initial horizontal velocity (vx) is given by:
vx = v1 * cos(theta1)
2. Vertical Component:
The initial vertical velocity (vy) is given by:
vy = v1 * sin(theta1)
Next, we need to find the equations for the motion of the projectile in the horizontal and vertical directions.
1. Horizontal Motion:
In the absence of drag, the only force acting on the projectile is due to gravity, mg. Therefore, the horizontal motion remains unchanged, and the equation for horizontal position (x) as a function of time (t) is given by:
x = vx * t
2. Vertical Motion:
In the presence of gravity and drag, the vertical motion is affected. The equations for vertical position (y) as a function of time (t), vertical velocity (vy), and acceleration (ay) are as follows:
y = vy * t + (1/2) * ay * t^2
vy = vy0 + ay * t
ay = -mg - (bv * vy) / m
b) To calculate the horizontal position of the peak of the trajectory (xpeak) and the height of the peak (ypeak), we need to find the time at which the vertical velocity becomes zero (vy = 0).
1. Find the time of flight (T) when vy = 0:
vy = vy0 + ay * t
0 = vy0 + ay * T
T = -vy0 / ay
2. Substituting T into the equation for y:
y = vy * t + (1/2) * ay * t^2
0 = vy0 * T + (1/2) * ay * T^2
ypeak = vy0 * T + (1/2) * ay * T^2
3. Substituting T into the equation for x:
x = vx * t
xpeak = vx * T
Note: To obtain precise values for xpeak and ypeak, the specific values of v1, theta1, m, g, and b must be known.
To analyze the flight of a projectile under the influence of gravity and linear drag, we need to consider the forces acting on the projectile and solve the equations of motion. Let's break it down step by step:
a. Determining the velocity components and coordinates as functions of time:
1. Break the initial velocity into its horizontal (v1x) and vertical (v1y) components:
v1x = v1 * cos(theta1)
v1y = v1 * sin(theta1)
2. Apply the equations of motion separately for horizontal and vertical directions:
Horizontal direction:
- The only force acting on the projectile horizontally is the drag force (fD = -bv). So the equation of motion becomes:
ma = -bv (where m is the mass of the projectile)
a = dvx/dt (acceleration is the derivative with respect to time)
dvx/dt = -b/m * vx
Vertical direction:
- The forces acting on the projectile vertically are gravity (mg) and drag force (-bv). So the equation of motion becomes:
ma = -mg - bv
a = dvy/dt
dvy/dt = -g - b/m * vy
3. Solve these differential equations to get the functional dependence of vx(t), vy(t), x(t), and y(t) on time:
dvx/dt = -b/m * vx
dvy/dt = -g - b/m * vy
The solutions to these equations can be expressed as:
vx(t) = v1x * exp(-b*t/m)
vy(t) = (v1y + (m*g/b)) * exp(-b*t/m) - (m*g/b)
x(t) = (m*v1x/b) * (1 - exp(-b*t/m))
y(t) = (m/b) * (v1y + (m*g/b)) * (1 - exp(-b*t/m)) - (m*g/b)*t
b. Calculating the horizontal position of the peak (xpeak) and the height for the peak (ypeak):
1. The horizontal position of the peak can be found by setting vy(t) equal to zero and solving for t:
vy(t) = 0
(v1y + (m*g/b)) * exp(-b*t/m) - (m*g/b) = 0
exp(-b*t/m) = (m*g/b) / (v1y + (m*g/b))
-b*t/m = ln((m*g/b) / (v1y + (m*g/b)))
t = -(m/b) * ln((m*g/b) / (v1y + (m*g/b)))
Substitute this value of t into x(t) to find xpeak:
xpeak = (m*v1x/b) * (1 - exp(-b*t/m))
2. The height at the peak can be found by substituting the value of t into y(t):
ypeak = (m/b) * (v1y + (m*g/b)) * (1 - exp(-b*t/m)) - (m*g/b)*t
It's important to note that the results of this analysis are specific to the given conditions, including the specific form of the drag force. The warning provided indicates that we cannot directly apply results from drag-free or turbulent-flow drag problems without considering the specific form of linear drag.