A particle moves in a vertical plane under gravity and an air resistance whose magnitude per unit mass is k times the speed of the particle, where k is a positive constant. The coordinates of the particle at time t, with respect to a horizontal axis Ox and a vertically upward axis Oy, are (x,y).

Show that

d^2 x /dt^2 = -k dx/dt

d^2y/dt^2 = - k dy/dt - g ...... (*)

Given that a projectile with a terminal speed of 30.5 m/s is fired when t=0 from O with an initial speed of 30.5 m/s at an inclination of 45 degree.

(a)Solve (*) for x and y in term of t.
(b) Plot the trajectory of the projectile.
(c) Compare the path trajectory to that when there is no air resistance.

To solve the equations d^2x/dt^2 = -k dx/dt and d^2y/dt^2 = -k dy/dt - g, we need to integrate them.

(a) Solving for x and y in terms of t:

For the equation d^2x/dt^2 = -k dx/dt, we can rearrange it as:

d^2x/dt^2 + k dx/dt = 0

This is a second-order linear homogeneous ordinary differential equation. We can solve it by assuming a solution of the form x = e^(rt), where r is a constant. Substituting this into the equation gives:

r^2 e^(rt) + k r e^(rt) = 0

Dividing both sides by e^(rt) and factoring out an r gives:

r(r + k) = 0

So we have two possible values for r: r = 0 and r = -k.

For r = 0, the solution is x = Ae^(0t) = A, where A is a constant.

For r = -k, the solution is x = Be^(-kt), where B is a constant.

So the general solution for the equation d^2x/dt^2 = -k dx/dt is:

x = A + Be^(-kt), where A and B are constants.

Now let's solve the equation d^2y/dt^2 = -k dy/dt - g:

Similar to the previous equation, assuming a solution of the form y = e^(st), we get:

s^2 e^(st) + k s e^(st) + g = 0

Dividing both sides by e^(st) and factoring out an s gives:

s^2 + ks + g = 0

This is a quadratic equation in s, which can be solved using the quadratic formula:

s = (-k ± sqrt(k^2 - 4g))/2

We have two possible values for s, which we'll call s1 and s2.

The general solution for the equation d^2y/dt^2 = -k dy/dt - g is then:

y = C1 e^(s1t) + C2 e^(s2t), where C1 and C2 are constants.

(b) To plot the trajectory of the projectile, we need to eliminate the constants A, B, C1, and C2. We can do this by substituting the given initial conditions.

When t = 0, the projectile is launched from O, so we have x(0) = 0 and y(0) = 0. We also know that the projectile has an initial speed of 30.5 m/s at an inclination of 45 degrees.

From the equation x = A + Be^(-kt), we can substitute x(0) = 0 to get:

0 = A + B

Similarly, from the equation y = C1 e^(s1t) + C2 e^(s2t), we can substitute y(0) = 0 to get:

0 = C1 + C2

We also have the initial speed of 30.5 m/s at an inclination of 45 degrees. This means that the initial horizontal velocity is v0 * cos(45 degrees) = (30.5 m/s) * cos(45 degrees), and the initial vertical velocity is v0 * sin(45 degrees) = (30.5 m/s) * sin(45 degrees).

The initial horizontal velocity is given by dx/dt, so we have dx/dt = (30.5 m/s) * cos(45 degrees).

The initial vertical velocity is given by dy/dt, so we have dy/dt = (30.5 m/s) * sin(45 degrees).

Substituting these initial velocities into the equations:

dx/dt = (30.5 m/s) * cos(45 degrees)
dy/dt = (30.5 m/s) * sin(45 degrees)

we get the following initial conditions:

x(0) = 0, y(0) = 0
dx/dt = (30.5 m/s) * cos(45 degrees)
dy/dt = (30.5 m/s) * sin(45 degrees)

By substituting these initial conditions into the general solutions for x and y, we can determine the values of the constants A, B, C1, and C2.

(c) To compare the path trajectory to that when there is no air resistance, we can solve the equations of motion without air resistance, which will be simpler.

To show that the given equations hold, we can use Newton's Second Law of Motion.

For the particle in the vertical plane, the gravitational force acting on the particle is -mg, where m is the mass of the particle and g is the acceleration due to gravity. The air resistance force acting on the particle is -kv, where v is the speed of the particle.

Using Newton's Second Law, we have:

ma_x = -kv_x ---(1)
ma_y = -mg - kv_y ---(2)

Let's differentiate equation (1) with respect to time to find the second derivative of x:

d^2x/dt^2 = (-d(kv_x)/dt) / m

Since v_x = dx/dt, we can rewrite the equation as:

d^2x/dt^2 = -k dv_x/dt

But dv_x/dt is the derivative of velocity with respect to time, which is acceleration. So, we can further simplify it to:

d^2x/dt^2 = -k dx/dt

This proves the first equation: d^2x/dt^2 = -k dx/dt

Now let's differentiate equation (2) with respect to time to find the second derivative of y:

d^2y/dt^2 = (-d(mg + kv_y)/dt) / m
= (-d(kv_y)/dt) / m
= -k dv_y/dt / m

Re-arranging the terms, we have:

m(d^2y/dt^2) + k(dy/dt) = -mg

Since dy/dt is the derivative of y velocity with respect to time, we can rewrite the equation as:

m(d^2y/dt^2) + k(v_y) = -mg

But v_y is the vertical component of velocity, it can be written as dy/dt. Therefore, the equation becomes:

m(d^2y/dt^2) + k(dy/dt) = -mg

This proves the second equation: d^2y/dt^2 = -k dy/dt - g

Now, let's solve the equations for x and y in terms of t:

Taking the integral of both sides of d^2x/dt^2 = -k dx/dt twice with respect to t, we get:

x = -kt + C1t + C2

Here, C1 and C2 are constants of integration.

Similarly, taking the integral of both sides of d^2y/dt^2 = -k dy/dt - g twice with respect to t, we get:

y = -kt - gt + C3t + C4

Here, C3 and C4 are constants of integration.

This gives us the solutions for x and y in terms of t.

To plot the trajectory of the projectile, substitute the values of x and y in terms of t into the equations:

x = -kt + C1t + C2
y = -kt - gt + C3t + C4

Now, compare the path trajectory with and without air resistance. When there is no air resistance, the particle follows a parabolic trajectory. However, with air resistance, the path trajectory will be affected. The particle will experience a gradual decrease in speed due to the air resistance and will not follow a perfect parabolic path. The air resistance force will cause the particle to fall faster and reach the ground sooner compared to the case when there is no air resistance.

In summary:
(a) The solutions for x and y in terms of t are:
x = -kt + C1t + C2
y = -kt - gt + C3t + C4

(b) To plot the trajectory, substitute the values of x and y in terms of t into the equations and plot the resulting curve.

(c) The path trajectory with air resistance will be different from the trajectory without air resistance. Air resistance will cause the particle to fall faster and reach the ground sooner, affecting the overall shape of the trajectory.