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

Explanation:

(A) Sum of forces on the projectile in the y direction:

-bv − mg = ma

Acceleration is the derivative of velocity with respect to time:

-bv − mg = m dv/dt

Separate the variables:

bv + mg = -m dv/dt

-1/m dt = 1/(bv + mg) dv

-b/m dt = b/(bv + mg) dv

Integrate:

-b/m t |₀ᵗ = ln(bv + mg) |₀ᵛ

-b/m (t − 0) = ln(bv + mg) − ln(0 + mg)

-b/m t = ln(bv + mg) − ln(mg)

-b/m t = ln((bv + mg) / mg)

e^(-b/m t) = (bv + mg) / mg

bv + mg = mg e^(-b/m t)

bv = -mg + mg e^(-b/m t)

v = -mg/b (1 − e^(-b/m t))

Velocity is derivative of position with respect to time:

dz/dt = -mg/b (1 − e^(-b/m t))

Separate the variables:

-b/(mg) dz = (1 − e^(-b/m t)) dt

Integrate:

-b/(mg) z |ᵧᶻ = (t + m/b e^(-b/m t)) |₀ᵗ

-b/(mg) (z − h) = (t + m/b e^(-b/m t)) − (0 + m/b e^(0))

-b/(mg) (z − h) = t + m/b e^(-b/m t) − m/b

z − h = -mg/b (t + m/b e^(-b/m t) − m/b)

z = h − mg/b (t + m/b e^(-b/m t) − m/b)

(B) Repeat steps from part A, but this time in the x direction.

-bv = ma

-bv = m dv/dt

-b/m dt = 1/v dv

-b/m t |₀ᵗ = ln v |ᵥᵛ

-b/m (t − 0) = ln vₓ − ln v₀ₓ

-b/m t = ln (vₓ / v₀ₓ)

vₓ / v₀ₓ = e^(-b/m t)

vₓ = v₀ₓ e^(-b/m t)

dx/dt = v₀ₓ e^(-b/m t)

dx = v₀ₓ e^(-b/m t) dt

x |₀ˣ = -m/b v₀ₓ e^(-b/m t) |₀ᵗ

x − 0 = -m/b v₀ₓ e^(-b/m t) − (-m/b v₀ₓ e^(0))

x = -m/b v₀ₓ e^(-b/m t) + m/b v₀ₓ

x = m/b v₀ₓ (1 − e^(-b/m t))

To find z(x), find t in terms of x then substitute into z(t).

b x / (m v₀ₓ) =  1 − e^(-b/m t)

e^(-b/m t) = 1 − b x / (m v₀ₓ)

-b/m t = ln(1 − b x / (m v₀ₓ))

t = -m/b ln(1 − b x / (m v₀ₓ))

z = h − mg/b (-m/b ln(1 − b x / (m v₀ₓ)) + m/b (1 − b x / (m v₀ₓ)) − m/b)

z = h − mg/b (-m/b ln(1 − b x / (m v₀ₓ)) + m/b − x / v₀ₓ − m/b)

z = h − mg/b (-m/b ln(1 − b x / (m v₀ₓ)) − x / v₀ₓ)

The range is when z = 0:

0 = h − mg/b (-m/b ln(1 − b x / (m v₀ₓ)) − x / v₀ₓ)

h = mg/b (-m/b ln(1 − b x / (m v₀ₓ)) − x / v₀ₓ)

bh/(mg) = -m/b ln(1 − b x / (m v₀ₓ)) − x / v₀ₓ

-(b/m)² h/g = ln(1 − (b/m) x / v₀ₓ) + (b/m) x / v₀ₓ

Unfortunately, this can't be simplified further without using something called the Lambert W function.

(C) The range of a projectile without air resistance launched horizontally from a height h is:

x = v₀ₓ √(2h/g)

no sign of dia of fuel cell

scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.