A catapult is designed to fire at a fixed angle of 45$^{o}$ above the horizontal (to maximize its range). A knight begins to charge from a distance of 100 m towards the catapult at a constant speed of 10 m/s. If the catapult is fired at the same instant as the knight begins to charge, what initial velocity must the stone fired from the catapult have to hit the knight?

Vertical displacement of stone $y=v_{0}\sin\theta t-\frac{1}{2}gt^2$ | Horizontal displacement of stone $x=v_{0}\cos\theta\, t$ |

At time of impact $t_{i}$

$0=v_{0}\sin\theta t_{i}-\frac{1}{2}gt_{i}^{2}$

$t_{i}=\frac{2v_{0}\sin\theta}{g}$

Horizontal displacement of knight: $100-10t$

$100-10t_{i}=v_{0}\cos\theta\, t_{i}$

$(v_{0}\cos\theta+10)\frac{2v_{0}\sin\theta}{g}=100$

$\frac{2v_{0}^{2}\sin\theta\cos\theta}{g}+\frac{20v_{0}\sin\theta}{g}=100$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}=100g$

$v_{0}^{2}+\frac{20}{\sqrt{2}}v_{0}+50=100g+50$

$(v_{0}+\frac{10}{\sqrt{2}})^2=100g+50$

$v_{0}=\frac{-10}{\sqrt{2}}\pm \sqrt{100g+50}$

$v_{0}=25.0 \mathrm{ms^{-1}}$ or $v_{0}=-39.2\mathrm{ms^{-1}}$ but we discard the negative solution.

Problem 3.60

Problem 3.61

Dealing with relative velocity is a particularly important application of addition and subtraction of vectors.

We can adopt a notation which can be helpful.

$\vec{v}_{AB}$: velocity of A relative to B

If we want to know the velocity of A relative to C if we know the velocity of A relative to B and the velocity of B relative to C

$\vec{v}_{AC}=\vec{v}_{AB}+\vec{v}_{BC}$

Whereas if we want to find the velocity of B relative to C

$\vec{v}_{BC}=\vec{v}_{AC}-\vec{v}_{AB}$

Here's a video that illustrates this concept in one dimension

Problem 3.58

Problem 3.62

A boat needs to get from a point $\vec{r_{1}}=(0\hat{i}+0\hat{j}) \,\mathrm{m}$ on one side of a river to a point $\vec{r_{2}}=(400\hat{i}+400\hat{j}) \,\mathrm{m}$ on the other side of the river. The river the boat is on has a velocity of $\vec{v_{R}}=(1\hat{i})\,\mathrm{m/s}$ relative to the river bank. The boat moves with constant velocity in a straight line from $\vec{r_{1}}$ to $\vec{r_{2}}$.

A. What is the angle between the path of the boat and the riverbank?

B. What is the **velocity** of the boat relative to the water (written in unit vector notation) if its speed relative to the water is 5 m/s?

C. What is the angle between the velocity vector of the boat relative to the water and the velocity vector of the water relative to the river bank?

A. $\tan\theta=\frac{400}{400}$ → $\theta=45^{o}$

B. $v_{BWx}^2+v_{BWy}^2=25$

$v_{BWx}+1=v_{BWy}$

This can either be solved by inspection (ie, if you see that its just a 3,4,5 triangle) or

$2v_{BWx}^2+2v_{BWx}+1=25$

$v_{BWx}^2+v_{BWx}-12=0$

$(v_{BWx}-3)(V_{BWx}+4)=0$

$v_{Bwx}=3$ (negative solution doesn't work) and it follows that $v_{Bwy}=4$

so $\vec{v}_{BW}=(3\hat{i}+4\hat{j})\mathrm{ms^{-1}}$

C. $\tan\phi=\frac{4}{3}=53.13^{o}$

$\vec{v}_{ce}=\vec{v}_{cr}+\vec{v}_{re}$

First convert the velocity of the canoe in to components along the direction of the river and perpendicular to it. This makes it a lot easier to subtract of the velocity of the river to find the velocity of the canoe relative to the river.

$\vec{v}_{se}=\vec{v}_{sr}+\vec{v}_{re}$

We need to draw the diagram using the known angles to set up the problem to find the speed the swimmer should maintain.

Relative to the water, the boat has a known velocity $v_{BW}$ (including direction $\theta_{BW}$).

The direction of the boat's velocity relative to the land can be deduced as $\tan\theta_{BL}=\frac{120\mathrm{m}}{280\mathrm{m}}$.

(Recall that $\tan\theta_{BL}=\frac{120\mathrm{m}}{280\mathrm{m}}$).