Linear motion and norms

Question

So I have the following question here...

Your friend will lauch a bottle rocket into the air from a launch pad. Upon launch, $0 \leq t \leq 7$ seconds, the rocket will have position

$$B(t)=(20t,30t,112t-16t^2)$$

given in feet, with a (north, east, up) orientation.

Your have built a tiny cannon that sits on the ground (so it shoots at a height of $0$ feet). It shoots pellets at a speed of $210$ feet per second. Your goal is to shoot a bottle $4$ seconds after its launch so that your pellet hits the bottle exactly one second after you shoot ($5$ seconds after the bottle is launched).

Given that you are at close range, you decide to model your shot using the linear motion

$$P(t)=(a,b,c)+(t-4)(d,e,f),$$

for $4\leq t\leq5$.

Note: The $(t-4)$ accounts for your shooting the pellet $4$ seconds after the bottle is launched, so $(a,b,c)$ is the location of your cannon (and hence your pellet) when you shoot it at $4$ seconds. The vector $(d,e,f)$ is the velocity of your pellet. Given the speed that the cannon shoots, we know that $||(d,e,f)||=210$.

a) At what position should your pellet be $5$ seconds after the bottle is launched? (i.e. What should $P(5)$ be?)

b) Given that you are shooting from the ground, what is $c$ and hence what must $f$ be?

c) You have to decide where to place your cannon to make the shot. If you set up your cannon so that is is somewhere on a line that is $40$ feet east of the bottle launch pad, then what will your linear motion function be?

d) Would any other position on the same line $40$ feet east of the launch pad work? Explain why or why not.

e) Use a 3D graphing program to graph $B(t)$ together with $P(t)$ and any other linear motions that you noticed in part d. Include a screenshot of your graph.

$\textbf{My attempt:}$

a) Would I not just plug in $P(5)$ here and get $(a+d,b+e,c+f)$? I don't think I should know what $a,b,c,d,e,f$ should be yet.

b) Since I am shooting form the ground, $c$ should be $0$ I believe. However, I am not sure what $f$ should be. I don't think it's $0$. The pellet would be in mid air by then so $f$ would not be $0$ and it should be something else but I am not sure what.

c) I want to say for this one, we take $B(t)$ and make it $B(t+40)$ and equate it with $P(t)$ somehow but I am not sure...

d) Not sure about this one.

e) This is fine.

If someone could help out with this question, that could be very nice.

Answer 1

Since you want to hit the bottle with the pellet at $t = 5$, then

$ P(5) = B(5) = (100, 150, 160 ) $

Now, using the approximation you made to the motion of the pellet,

$ P(t) = P_0 + (t - 4) P_1 $

where $P_0 = (a,b,c)$ and $P_1 = (d,e,f) $

And all we know is that $\| P_1 \| = 210 $

Therefore, we want

$ (100, 150, 160) = P_0 + P_1 $

At $t=4$ the pellet is still at the ground, so $c = 0$, also $f \gt 0$.

Looking at the above equation, one determines immediately that $f = 160$.

Now we have the following system of equations

$ a + d = 100 $

$ b + e = 150 $

and in addition to these two linear equations, we have the quadratic equation:

$ d^2 + e^2 + f^2 = 210^2 $

That is (since $f = 160$)

$ d^2 + e^2 = 210^2 - 160^2 = 18500 $

This equation suggests that we take

$ d = \sqrt{18500} \cos \phi $ and $ e = \sqrt{18500} \sin \phi $

That is,

$ d = 10 \sqrt{185} \cos \phi $ and $ e = 10 \sqrt{185} \sin \phi $

Then it would follows from the linear system that

$ a = 100 - 10 \sqrt{185} \cos \phi $

$ b = 150 - 10 \sqrt{185} \sin \phi $

Now from the specification in part (c), we now have to set $ b = 40 $, while $a$ is open.

From the last equation,

$ 40 = 150 - 10 \sqrt{185} \sin \phi $

So that

$ \sin \phi = \dfrac{ 11}{\sqrt{185}} = 0.8087$

There are two values of $\phi$ with this $\sin$. They differ by the sign of their $\cos$. So, we have

$ \cos \phi_1 = \sqrt{ 1 - 0.8087^2 } = 0.58817$

and

$ \cos \phi_2 = - \sqrt{1 - 0.8087^2 } = -0.58817 $

This answers part (d), namely there are two possible positions. with $\phi_1$, we have

$ a = 100 - 10 \sqrt{185} \cos \phi_1 = 20 $

$ b = 40 $

$ d = 100 -a = 80 $

$ e = 150 - b = 110 $

$ f = 160 $

With $ \phi_2 $

$ a = 100 + 80 = 180 $

$ b = 40 $

$ d = 100 - 180 = -80$

$ e = 150 - 40 = 110$

$ f = 160 $

Answer 2

We can proceed as follows.

a) At $t=5$ we need $P(5)=B(5)$, that is $(a+d,b+e,c+f)=(100,150,160)$

b) We are taking $P(t)$ such that at $t=4$ we have $c=0$ and therefore $f=160$.

c) We are taking $P(t)$ such that at $t=4$ we have $b=40$ and therefore $e=110$. Now we can find $d$ from

• $d^2+e^2+f^2=210$

and then $a$ for the complete solution.