Question

How High?—No Air Resistance Suppose a small cannonball weighing 16 pounds is shot vertically upward, as shown in Figure 3.1.13, with an initial velocity v0 5 300 ft/s. The answer to the question “How high does the cannonball go?” depends on whether we take air resistance into account.

(a) Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by d2 sydt2 5 2g (equation (12) of Section 1.3). Since dsydt 5 v(t) the last differential equation is the same as dvydt 5 2g, where we take g 5 32 ft/s2 . 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. Find the maximum height attained by the cannonball.

Short answer

(a) $v\left(t\right)=-32t+300$

(b) $s\left(t\right)=-16t^2+300t$ and $s=1406.25\text{ft}$

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$\mathit{L}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{i}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

a) We have a small cannonball with a weight Ib with a velocity is shorted vertically upward with no air resistance, has the differential equation

$\frac{dv}{dt}=-g$

with the initial condition

$v_0=300\text{ft}/\text{s}$

and we have to solve it as the following technique :

Since we havethen we can have$g=32\text{ft}/{\text{s}}^2,$ the differential equation shown in (1) as

$\frac{dv}{dt}=-32$

Separate the variables and integrate

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

$\begin{array}{rcl}dv& =& -32dt\\ \int dv& =& -32\int dt\\ v\left(t\right)& =& -32t+k\end{array}$

After that, to find the value of constant, we have to apply the point of condition$(v,t)=(300,0)$into equation (2), then we have

$300=-32\left(0\right)+k$

Then we have

$k=300$

After that, substitute with the value of constantinto equation (2), then we have

$v\left(t\right)=-32t+300$

is the velocity of the small cannonball at time t.

Find the distance

b) Now, we have the distance measured from releasing point to the position of cannonball at time is related to the velocity of the cannonball as

$\frac{ds}{dt}=v$

with the condition

$s(t=0)=0$

This differential equation (a) is a linear and separable, then we can obtain the distance at time t as

$\int ds=\int vdt$

Then we have

$\begin{array}{rcl}s& =& \int (-32t+300)dt\\ & =& \int -32tdt+\int 300dt\\ & =& -32\int tdt+300\int dt\\ & =& -32\left(dfrac{t}^{2}2\right)+300t+h\\ & =& -16t^2+300t+h\end{array}$

After that, to find the value of constantwe have to apply the point of condition$(s,t)=(0,0)$ into equation (a), then we have

$0=-16\left(0\right)+300\left(0\right)+h$

Then we have

$h=0$

Substitution

After that, substitute with the value of constant into equation (b), then we have

$s\left(t\right)=-16t^2+300t$

is the distance for the cannonball from the shotting point at time t.

Now, we have to obtain the maximum height that the cannonball reaches as the following:

First, we have to obtain the time at which the velocity of cannonball is equal to 0 by substituting with $v=0$into equation (3) as

$0=-32t+300$

Then we have

$\begin{array}{rcl}t& =& \frac{300}{32}\\ & =& \frac{75}{8}\text{seconds}\end{array}$

Second, substitute with $t=\frac{75}{8}$seconds into equation (c), then we have

$\begin{array}{rcl}s& =& -16{\left(\frac{75}{8}\right)}^2+300\left(\frac{75}{8}\right)\\ & =& -16\left(\frac{5,625}{64}\right)+\frac{5,625}{2}\\ & =& -\frac{5,625}{4}+\frac{5,625}{2}\\ & =& \frac{5,625}{4}\\ & =& 1406.25\text{ft}\end{array}$

a) $v\left(t\right)=-32t+300$

b) $s\left(t\right)=-16t^2+300t$and$s=1406.25\text{ft}$