Answer:
tm = tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] }
Xm = Xₐ = (v₀m)/k - ({m²g}/k²) ㏑(1+{kv₀/mg})
Step-by-step explanation:
Note, I substituted maximum time tm = tₐ and maximum height Xm = Xₐ
We will use linear ordinary differential equation (ODE) to solve this question.
Remember that Force F = ma in 2nd Newton law, where m is mass and a is acceleration
Acceleration a is also the rate of change in velocity per time. i.e a=dv/dt
Therefore F = m(dv/dt) = m (v₂-v₁)/t
There are two forces involved in this situation which are F₁ and F₂, where F₁ is the gravitational force and F₂ is the air resistance force.
Then, F = F₁ + F₂ = m (v₂-v₁)/t
F₁ + F₂ = -mg-kv = m (v₂-v₁)/t
Divide through by m to get
-g-(kv/m) = (v₂-v₁)/t
Let (v₂-v₁)/t be v¹
Therefore, -g-(kv/m) = v¹
-g = v¹ + (k/m)v --------------------------------------------------(i)
Equation (i) is a inhomogenous linear ordinary differential equation (ODE)
Therefore let A(t) = k/m and B(t) = -g --------------------------------(ia)
b = ∫Adt
Since A = k/m, then
b = ∫(k/m)dt
The integral will give us b = kt/m------------------------------------(ii)
The integrating factor will be eᵇ = e ⁽<em>k/m</em>⁾
The general solution of velocity at any given time is
v(t) = e⁻⁽b⁾ [ c + ∫Beᵇdt ] --------------------------------------(iiI)
substitute the values of b, eᵇ, and B into equation (iii)
v(t) = e⁻⁽kt/m⁾ [ c + ∫₋g e⁽kt/m⁾dt ]
Integrating and cancelling the bracket, we get
v(t) = ce⁻⁽kt/m⁾ + (e⁻⁽kt/m⁾ ∫₋g e⁽kt/m⁾dt ])
v(t) = ce⁻⁽kt/m⁾ - e⁻⁽kt/m⁾ ∫g e⁽kt/m⁾dt ]
v(t) = ce⁻⁽kt/m⁾ -mg/k -------------------------------------------------------(iv)
Note that at initial velocity v₀, time t is 0, therefore v₀ = v(t)
v₀ = V(t) = V(0)
substitute t = 0 in equation (iv)
v₀ = ce⁻⁽k0/m⁾ -mg/k
v₀ = c(1) -mg/k = c - mg/k
Therefore c = v₀ + mg/k ------------------------------------------------(v)
Substitute equation (v) into (iv)
v(t) = [v₀ + mg/k] e⁻⁽kt/m⁾ - mg/k ----------------------------------------(vi)
Now at maximum height Xₐ, the time will be tₐ
Now change V(t) as V(tₐ) and equate it to 0 to get the maximum time tₐ.
v(t) = v(tₐ) = [v₀ + mg/k] e⁻⁽ktₐ/m⁾ - mg/k = 0
to find tₐ from the equation,
[v₀ + mg/k] e⁻⁽ktₐ/m⁾ = mg/k
e⁻⁽ktₐ/m⁾ = {mg/k] / [v₀ + mg/k]
-ktₐ/m = ㏑{ [mg/k] / [v₀ + mg/k] }
-ktₐ = m ㏑{ [mg/k] / [v₀ + mg/k] }
tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] }
Therefore tₐ = -m/k ㏑{ [mg/k] / [v₀ + mg/k] } ----------------------------------(A)
we can also write equ (A) as tₐ = m/k ㏑{ [mg/k] [v₀ + mg/k] } due to the negative sign coming together with the In sign.
Now to find the maximum height Xₐ, the equation must be written in terms of v and x.
This means dv/dt = v(dv/dx) ---------------------------------------(vii)
Remember equation (i) above -g = v¹ + (k/m)v
Given that dv/dt = v¹
and -g-(kv/m) = v¹
Therefore subt v¹ into equ (vii) above to get
-g-(kv/m) = v(dv/dx)
Divide through by v to get
[-g-(kv/m)] / v = dv / dx -----------------------------------------------(viii)
Expand the LEFT hand size more to get
[-g-(kv/m)] / v = - (k/m) / [1 - { mg/k) / (mg/k + v) } ] ---------------------(ix)
Now substitute equ (ix) in equ (viii)
- (k/m) / [1 - { mg/k) / (mg/k + v) } ] = dv / dx
Cross-multify the equation to get
- (k/m) dx = [1 - { mg/k) / (mg/k + v) } ] dv --------------------------------(x)
Remember that at maximum height, t = 0, then x = 0
t = tₐ and X = Xₐ
Then integrate the left and right side of equation (x) from v₀ to 0 and 0 to Xₐ respectively to get:
-v₀ + (mg/k) ㏑v₀ = - {k/m} Xₐ
Divide through by - {k/m} to get
Xₐ = -v₀ + (mg/k) ㏑v₀ / (- {k/m})
Xₐ = {m/k}v₀ - {m²g}/k² ㏑(1+{kv₀/mg})
Therefore Xₐ = (v₀m)/k - ({m²g}/k²) ㏑(1+{kv₀/mg}) ---------------------------(B)
and the radius, r = 50 cm, you can plug those values into the equation for the lateral area and solve for l, the slant height:
