Question

We consider a projectile motion against a linear drag force D = −b∗v, where v is the velocity

Answer 1

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)