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)

answer set (b) 0.5, 3/4 80% is in order from least to greatest.  

step-by-step explanation:

in order to compare rational numbers to each other, it is much easier if they are in the same form.   typically, the best form in which to evaluate rational numbers is in decimals.   in order to convert a percent to a decimal, you would divide the given percentage by 100.   you can also accomplish the conversion by moving the decimal point over two places to the left.   for example, the decimal value of 80% is 80/100, or 0.8.   in order to convert a fraction to a decimal, you need to divide the numerator by the denominator.   for example, 3/4 is 0.75.   so, for choice (b) 0.5 3/4 (0.75), 80% (0.8) is the only choice in which the numbers are in order from least to greatest.  

add the numbers under frequency to find the total number:

5 + 8 +7 + 4 + 0 +1 = 25

the answer is 25.