Abody of constant mass m is projected vertically upward with an initial velocity v0 in a medium offering a resistance k|v|, where k is a constant. neglect changes in the gravitational force. a. find the maximum height xm attained by the body and the time tm at which this maximum height is reached.

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 ⁽k/m⁾

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)

the first one, all those points on the line just make a strait line.

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!