Answer to Question #122576 in Differential Equations for Jse

Question

Suppose a small cannonball weighing 98 N is shot vertically upward, with an initial velocity
v0 = 82 m/s.If air resistance is ignored and the positive direction is upward, then a model for the state of the cannonball is given by d^2s/dt^2 = −g. Since ds/dt = v(t)the last differential equation is the same as dv/dt = −g, where we take g = 9.8 m/s2.If air resistance is incorporated into the model, it stands to reason that the maximum height attained by the cannonball must be less than if air

A. If the positive direction is upward, a model for the state of the cannonball is given by
m dv/dt = −mg − kv, where m is the mass of the cannonball and k > 0 is a constant of proportionality. Suppose k = 0.0025 and find the velocity v(t) of the cannonball at time t.

B. Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level.

C. Find the maximum height attained by the cannonball.

Answer 1

Let mass, height, velocity, time are respectively represented by m, v, s , t.

So mg = 98N where g = 9.8 m/s²

So m = 98/9.8 = 10 kg

Initial velocity v₀ = 82 m/s

(A)

A model of the state of the cannon ball is given by

$m\frac{dv}{dt}$ = - mg - kv

=> m$\frac{dv}{dt}$ = -k(v + $\frac{mg}{k}$ )

=> $\frac{dv}{(v+\frac{mg}{k})}$ = $-\frac{k}{m} dt$

Integrating,

$\int$ $\frac{dv}{(v+\frac{mg}{k})}$ = $\int$ $-\frac{k}{m} dt$

=> ln$(v+\frac{mg}{k})$ = $-\frac{k}{m}t$ + C

By initial condition, when t=0, v= 82

So ln$(82+\frac{mg}{k})$ = C

So ln$(v+\frac{mg}{k})$ = $-\frac{k}{m}t$ + ln$(82+\frac{mg}{k})$

=> ln$(v+\frac{mg}{k})$ - ln$(82+\frac{mg}{k})$ = $-\frac{k}{m}t$

=> ln $\frac{(v+\frac{mg}{k})}{(82+\frac{mg}{k})}$ = $-\frac{k}{m}t$

=> $\frac{(v+\frac{mg}{k})}{(82+\frac{mg}{k})}$ = $e^{-\frac{k}{m}t}$

=> $(v+\frac{mg}{k})$ = $(82+\frac{mg}{k})$ $e^{-\frac{k}{m}t}$

=> $v = -\frac{mg}{k}$ + $(82+\frac{mg}{k})$ $e^{-\frac{k}{m}t}$

Putting the values of m,k,g we get

$\frac{mg}{k} = \frac{9.8*10}{0.0025}=39200$

$\frac{k}{m} = 0.00025$

So expression for velocity is

v(t) = -39200+39282$e^{-0.00025t}$

(B)

From part - A we can write

$\frac{ds}{dt}$ = -39200+39282$e^{-0.00025t}$

=> ds = (-39200 + 39282$e^{-0.00025t}$ )dt

$\int$ ds = $\int$(-39200 + 39282$e^{-0.00025t}$ )dt

=> s = -39200t - $\frac{39282}{0.00025}$ $e^{-0.00025t}$ + C₁

By initial condition, s = 0 when t=0

C₁ = $\frac{39282}{0.00025} = 157128000$

So s = 157128000 - 39200t - 157128000$e^{-0.00025t}$

So height of the cannon ball is

s(t)=157128000(1- $e^{-0.00025t}$ ) - 39200t

(C)

For maximum height $,\frac{ds}{dt} = v = 0$

=> -39200 + 39282$e^{-0.00025t} = 0$

=> $e^{-0.00025t} = \frac{39200}{39282}$

=> -0.00025t = ln($\frac{39200}{39282}$ )

=> -0.00025t = - 0.00208965189

=> t = $\frac{0.00208965189}{0.00025}$ = 8.35860756

Putting this value of t in s(t) we get maximum height.

So maximum height is

157128000(1 - $e^{-0.00025*8.35860756}$ ) - 39200*8.35860756

= 328000-327657.416

= 342.584

So maximum height is 342.584 m