Question

What goes up: (a) It is well known that the model in which air resistance is ignored, part (a) of Problem 36, predicts that the time ${\mathit{t}}_{\mathbf{\alpha }}$it takes the cannonball to attain its maximum height is the same as the time ${\mathit{t}}_{\mathbf{d}}$it takes the cannonball to fall from the maximum height to the ground. Moreover, the magnitude of the impact velocity ${\mathit{v}}_{\mathbf{1}}$will be the same as the initial velocity ${\mathit{v}}_{\mathbf{0}}$of the cannonball. Verify both of these results.

(b) Then, using the model in Problem 37 that takes air resistance into account, compare the value of ${\mathit{t}}_{\mathbf{\alpha }}$with ${\mathit{t}}_{\mathbf{d}}$and the value of the magnitude of ${\mathit{v}}_{\mathbf{1}}$and ${\mathit{v}}_{\mathbf{0}}$. A root-finding application of a CAS (or graphic calculator) may be useful here.

Short answer

a) $t_d=9.375$seconds and $v_0=v_i=300ft/s$but in opposite direction

b) $t_d>t_a$ and $v_i<v_0$

Step-by-step solution

Find the cannonball td

a) From the solution of problem (36), we have the velocity of the cannonball at time t as

$v\left(t\right)=-32t+300$---------- (1)

Now, to find the time needed for the cannonball to attain its maximum height, we have to substitute by $v=0$into equation (1), then we have

$\begin{array}{rcl}0& =& -32t_a+300\\ 32t_a& =& 300\end{array}$

Then we have

$\begin{array}{rcl}{t}_a& =& \frac{300}{32}\end{array}$

$\begin{array}{rcl}{t}_a& =& 9.375\end{array}$seconds, is the time which the cannonball reaches its maximum height.

After that, also from the solution of problem (36), we have the height of the cannonball at time t as

$\begin{array}{rcl}s\left(t\right)& =& -16t^2+300t\end{array}$----------- (2)

Now, to find the time needed for the cannonball takes to reach ground from its shooting by substitute by $\begin{array}{rcl}s& =& 0\end{array}$into equation (2), then we have

$\begin{array}{rcl}0& =& -16t^2+300t\\ 16t& =& 300\end{array}$

Then we have

$\begin{array}{rcl}t& =& \frac{300}{16}\end{array}$

Seconds

Finally, we obtain the time $\begin{array}{rcl}& & t_d\end{array}$at which the cannonball reaches the ground from its maximum height as

$\begin{array}{rcl}{t}_d& =& t-t_a\\ & =& 18.75-9.375\\ & & 9.375sec\end{array}$

Which is the same as the time$\begin{array}{rcl}& & t_d\end{array}$ at which the cannonball reaches its maximum height.

Solution of initial velocity of the cannonball

Now, we can obtain the value of the initial velocity of the cannonball $v_0$by substitute with $t=0$into equation (1) as

$\begin{array}{rcl}{v}_0& =& -32\left(0\right)+300\\ & =& 300ft/s\end{array}$

Also we can obtain the impact velocity of thecannonball$\begin{array}{rcl}& & v_i\end{array}$ bysubstitutingwith$\begin{array}{rcl}t& =& 18.75\end{array}$ into equation (1) as

$\begin{array}{rcl}{v}_i& =& -32\times \left(18.75\right)+300\\ & =& -600+300\\ & =& -300ft/s\end{array}$

Is the same as the initial velocity but in opposite directions.

Solution of problem (37)

b) From the solution of problem (37), we have the velocity of the cannonball at time t as

$v\left(t\right)=6700e^{-\frac{1}{200}t}-6400$----- (3)

Now, to find the time needed for the cannonball to attain its maximum height, we have to substitute by $v=0$into equation (3), then we have

$\begin{array}{rcl}0& =& 6700e^{-\frac{1}{200}t}-6400\\ e^{-\frac{1}{200}t}& =& 0.9552\\ -\frac{1}{200}t& =& -0.04581\end{array}$

Then we have

$\begin{array}{rcl}{t}_a& =& 0.04581\times 200\\ & =& 9.162sec\end{array}$

is the time which the cannonball reaches its maximum height.

After that, also from the solution of problem (36), we have the height of the cannonball at time t as

$\begin{array}{rcl}s\left(t\right)& =& -6400t-1340000e^{\frac{1}{200}t}+1340000\end{array}$--------- (4)

Now, to find the time needed for the cannonball takes to reach ground from its shooting by $\begin{array}{rcl}s& =& 0\end{array}$substitute by into equation (4), then we have

role="math" localid="1664361738288" $\begin{array}{rcl}0& =& -6400t-1340000e^{-\frac{1}{200}t}+1340000\\ t+209.375e^{-\frac{1}{200}t}& =& 209.375\end{array}$

Then we have

$\begin{array}{rcl}t& \approx & 18.466\end{array}$seconds

Finally, we obtain the time $\begin{array}{rcl}& & t_d\end{array}$at which the cannonball reaches the ground from its maximum height as

$\begin{array}{rcl}{t}_d& =& t-t_a\\ & =& 18.466-9.162\end{array}$

$\begin{array}{rcl}& =& 9.304\end{array}$seconds, is greater that the time$\begin{array}{rcl}& & t_d\end{array}$ at which the cannonball reaches its maximum height.

Solution of initial velocity of the cannonball

Now, we can obtain the value of the initial velocity of the cannonball $v_0$by substitute with $t=0$into equation (3) as

$\begin{array}{rcl}{v}_0& =& 6700e^0-6400\\ & =& 300ft/s\end{array}$

Also we can obtain the impact velocity of thecannonball bysubstitutingwith into equation (3) as

$\begin{array}{rcl}{v}_i& =& 6700e^{-\frac{1}{200}\times 18.466}-6400\\ & =& 6700\times 0.9118-6400\\ & =& 6190.1-6400\\ & =& -290.9ft/s\end{array}$

, which is lower that the initial velocity but in opposite direction.