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)

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong