Castle siege

(This has nothing to do with the solution but I find this problem hilarious!)

Okay, notice that the displacement of the boulder along the $y$ -axis (upwards) is equal to $(10+1.75)$ $\text{m}=11.75$ $\text{m}$

And the velocity along the $y$ -axis is equal to $v\sin\theta$ $\text{m/s}=50\sin\theta$ $\text{m/s}$ .

Let's say that the time it takes for the boulder to hit the guard is $t$ .

So, $11.75=50\sin\theta t +\frac{1}{2} g t^2$ [ $g$ is the acceleration of gravity].

Or $11.75=50\sin\theta t - 4.9 t^2$ $\cdots(1)$ .

The displacement along the $x$ - axis (horizontally) is equal to $100$ $\text{m}$ .

And the velocity along the $x$ -axis is equal to $v \cos \theta$ $\text{m/s}$ .

So we have another equation:

$100=50\cos \theta t$ [ $t$ is the same $t$ as before]

In other words, $t=\frac{100}{50\cos\theta}=\frac{2}{\cos\theta}$ .

Let's plug this value of $t$ in $(1)$ .

This is the equation we get:

$11.75=100\tan \theta-19.6\sec^2\theta$

$\Rightarrow 11.75=100\tan \theta-19.6(1+\tan^2\theta)$

And after a little bit of re-arranging-

$19.6\tan^2\theta-100\tan\theta+31.35=0$ .

Solving this for $\tan\theta$ gives us two values of $\tan\theta$ which in turn gives us two values of $\theta$ : $18.55^\circ$ and $78.15^\circ$ .

Both of them happen to be correct although I'm not sure if Brilliant accepts both the answers (you can't submit any more answers after getting the problem right).

The first value of $\theta$ implies that the boulder hit the guard before attaining maximum height. The second one means the boulder attained maximum height and as it was falling down(and moving forward at the same time), it hit the guard.

The figure resembles the first scenario. So that's the one I went with.

So, (in my case) $\theta=18.55$ $\text{degrees}$ .

Let us establish an expression for the height of the boulder $h$ in terms of the gravitational acceleration $g$ , the horizontal distance $x$ , the initial speed $v$ and the angle $\theta$ , and then solve for $\theta$ .

We can decompose the initial velocity of the boulder in a horizontal component $v_x = v \cos\theta$ and a vertical component $v_y = v \sin\theta$ . Since gravity points straight down, the horizontal component of the velocity will remain constant. The time it takes the boulder to cover the horizontal distance $x$ is thus

$t = \frac{x}{v_x} = \frac{x}{v \cos\theta}.$

The height of the boulder at time $t$ , given an initial height $0$ , initial velocity $v_y$ and constant acceleration $g$ is

$h = v_y t + \frac12 g t^2 = v \sin\theta \frac{x}{v \cos\theta} + \frac12 g \frac{x^2}{v^2 \cos^2\theta} \\ \quad = x \tan\theta + \frac{gx^2}{2v^2} (1 + \tan^2\theta),$

where I used the identity $\frac1{\cos^2\theta} = 1 + \tan^2\theta$ . Since we wish to solve for $\theta$ , we recognize a second-degree equation in $\tan\theta$ :

$0 = \frac{gx^2}{2v^2} \tan^2\theta + x \tan\theta + \frac{gx^2}{2v^2} - h,$

with the solution(s)

$\tan\theta = \frac{v^2}{gx^2} \left( -x \pm \sqrt{x^2 - \frac{2gx^2}{v^2} \left( \frac{gx^2}{2v^2} - h \right) } \right) \\ \qquad = \frac1{gx} \left( -v^2 \pm \sqrt{v^4 + 2ghv^2 - g^2x^2} \right).$

When we plug in the numbers $g = -9.8\,\mathrm{m/s^2}$ , $h = 11.75\,\mathrm{m}$ , $x = 100\,\mathrm{m}$ and $v = 50\,\mathrm{m/s}$ , we find two possible solutions for $\theta$ . In the spirit of the problem statement, we will take the solution with the smallest value of $\theta$ , which gives the guard the least amount of time to react:

$\theta = \boxed{18.55^\circ}$

The other solution is $\theta = 78.15^\circ$ .

From the tarjectory of a particle undergoing projectile motion we have

$y=x\tan\theta-\displaystyle\frac{gx^2\sec^2\theta}{2u^2}$ $\,----(*)$

where $x,y$ are co-ordinates of particle w.r.t launching point, $\theta$ is launch angle w.r.t horizontal, $g$ is acceleration due to gravity, $u$ is launching speed.

Given the data we have $y=10+1.75=11.75\,m$ , $x=100\,m$ , $g=9.8\,ms^{-2}$ , $u=50\,ms^{-1}$ . We need to find $\theta$ .

Substituting the values in $(*)$ we have

$11.75=100\tan\theta-\displaystyle\frac{9.8\times100^2\times(\tan^2\theta+1)}{2\times50^2}$

$\Rightarrow$ $(2\times9.8)\times\tan^2\theta-(100)\times\tan\theta+(11.75+2\times9.8)=0$

Solving for $\theta$ gives $\tan\theta=\displaystyle\frac{100\pm\sqrt{100^2-94\times9.8-16\times9.8^2}}{4\times9.8}$

which yields $\theta=18.55^\circ,\,78.15^\circ$ both of which satisfy the given conditions.

We start by defining the desired height $H=11.75m$ and the length $L=100m$ .

The starting velocity $v_{0}=50m/s$ is decomposed into the horizontal direction ( $x$ ) and the vertical direction ( $y$ ) according to the starting angle $\theta$ :

$v_{0x}=v_{0} cos \theta$

$v_{0y}=v_{0} sin \theta$ .

Gravity $g$ only acts on the vertical velocity, and so the velocities as functions of time $t$ become:

$v_{x} (t) = v_{0x} = v_{0} cos \theta$

$v_{y} (t) = v_{0y} - g t = v_{0} sin \theta - gt$ .

Integrating these functions with respect to $t$ gives the distance functions:

$D_{x} (t) = v_{0} cos \theta \cdot t$

$D_{y} (t) = v_{0} sin \theta \cdot t - \frac {1}{2} g t^{2}$ .

When the boulder reaches the position $D_{x} = L$ , the time $T$ will be:

$T= \frac {L}{v_{0} cos \theta }$ ,

and this is exactly the time when the vertical position must be $D_{y} = H$ :

$H = v_{0} sin \theta \cdot T - \frac {1}{2} g T^{2}$

or

$H = v_{0} sin \theta \cdot \frac {L}{v_{0} cos \theta } - \frac {1}{2} g ( \frac {L}{v_{0} cos \theta } ) ^{2}$ .

By rearrangement, this becomes:

$H cos ^2 \theta - L sin \theta \cdot cos \theta = - \frac{g L^2}{2 v_{0}^2}$ .

We now introduce the substitutions

$cos ^2 \theta = \frac {1}{2} + \frac {1}{2} cos 2 \theta$

$sin \theta \cdot cos \theta = \frac {1}{2} sin 2 \theta$ .

which transform the equation into:

$H cos 2 \theta - L sin 2 \theta = - ( \frac {gL^2}{v_{0}^2} + H )$ .

In order to solve this equation, we can further transform it by exploiting the general equivalence

$a cos \alpha + b sin \alpha = c$

$\Leftrightarrow$

$R cos (\alpha - \phi ) = c$ , where $R= \sqrt {a^2 +b^2}$ and $tan \phi = \frac {b}{c}$ .

For our equation this yields:

$R= \sqrt {H^2 +L^2}$

and

$tan \phi = - \frac{L}{H} \Rightarrow \phi = -83.29 ^ \circ$ :

$cos (2 \theta - \phi ) = - ( \frac {gL^2}{v_{0}^2} + H ) \cdot \frac {1}{ \sqrt {H^2 + L^2 } }$ .

Finally now, inserting the numerical values of the variables, the solution becomes:

$2 \theta + 83.29^ \circ = 120.4^ \circ$

$\boxed { \theta = 18.55^ \circ }$ .

From the law of projectile we know that,

$y = (\tan \theta) x - \frac{g x^{2}}{2 (v\cos \theta)^{2}}$

or, $\frac{y}{x} = \tan \theta - \frac{gx}{2v^{2}(\cos \theta)^{2}}$

where, y= height of the projectile = (10+1.75)m=11.75m

x= distance travelled by the projectile = 100m

g = gravitational acceleration = 9.8 m\s^2

$\theta$ = angle to launch the boulder

v = the velocity = 50 m\s

so from the equation above we get

$\frac{y}{x} = \tan \theta - \frac{gx}{2v^{2}} \times \frac{1}{(\cos \theta)^{2}}$

Let, $\frac{y}{x} = a$ and $\frac{gx}{2v^{2}}=b$

Now the equation becomes,

$a = \tan\theta - b\times \frac{1}{(\cos\theta)^{2}}$

or, $a = \tan\theta - b(\sec\theta)^2$

or, $a = \tan\theta - b(1+(\tan\theta)^2)$ or, $a = \tan\theta - b - b(\tan\theta)^{2}$

or, $b(\tan\theta)^{2} - \tan\theta + (a+b) = 0$ And with value of $a$ and $b$

$0.196 \times(\tan\theta)^{2} - \tan\theta + 0.3135$

By solving the equation we get,

$\tan\theta = 4.76646$ and $tan\theta = 0.3355$ so, $\theta = 78.15$ or $\theta = 18.55$

The first Value doesn't satisfy our equation.

Hence,

$\theta = 18.55$

$v_{x} = 50 \cos \theta$ is the horizontal component of the velocity.

$v_{y} = 50 \sin \theta$ is the vertical component of the velocity.

$h = 11.75m$ is the target height.

$t_{hit}$ is the flight time before hitting the target.

In the horizontal direction we get $v_{x}t_{hit} = 100$ , so $t_{hit} = 2 \sec \theta$

In the vertical direction: $\begin{aligned} h(t) &= 50t \sin \theta - gt^2/2\ \\ h(t_{hit}) &= 100 \tan \theta - 2g \sec^2 \theta \\ &= 100 \tan \theta - 2g - 2g \tan^2 \theta \\ 0 &= 2g \tan^2 \theta - 100 \tan \theta + (2g + 11.75) \end{aligned}$ Using the quadratic formula, $\tan \theta = \frac{100 \pm \sqrt{10000 - 8g(2g + 11.75)}}{4g}$

Putting this into the calculator and choosing the sign that gives a sensible answer gives us $\tan \theta = 0.3355711...$ so $\theta = 18.55...{}^{\circ}$

When dealing with this projectile question you have to component the initial velocity into its vertical ( $50 \sin \theta$ ) and horizontal ( $50 \cos \theta$ ) components and deal with them separately:

In the horizontal direction there is no acceleration so there is a constant velocity so you can use $speed = \frac{distance}{time}$ The only unknown in this equation is the time and so you end up with the time taken for the projectile flight to be $\frac{100}{50 \cos \theta}$ .

In the vertical direction there is a constant acceleration on the projectile of $-9.8ms^{-2}$ which means that you have to use the equations of constant acceleration (the "suvat" equations). In this scenario, the most appropriate one to use is: $s = ut + \frac{1}{2} at^{2}$ You know that the vertical displacement, s, is $11.75m$ the initial vertical velocity, u, is $50 \sin \theta ms^{-1}$ , the time taken, t, is $\frac{100}{50 \cos \theta} s$ and the acceleration, a, is $-9.8ms^{-2}$ . When you substitute this into the equation, you get:

$11.75 = 50 \sin \theta . \frac{100}{50 \cos \theta} + \frac{1}{2} . -9.8 . (\frac{100}{50 \cos \theta})^{2}$

This equation can be simplified by using the trig identity: $\tan \theta = \frac{ \sin \theta}{ \cos \theta}$

$11.75 = 100 \tan \theta - \frac{19.6}{ \cos^{2} \theta }$

We know that $\frac{1}{ \cos \theta} = \sec \theta$ :

$11.75 = 100 \tan \theta - 19.6 \sec^{2} \theta$

We now need to use another trig identity: $\sec^{2} \theta = 1 + \tan^{2} \theta$

$11.75 = 100 \tan \theta - 19.6(1 + \tan^{2} \theta )$

From which you can form a quadratic equation in terms of $\tan \theta$ :

$19.6 \tan^{2} \theta - 100 \tan \theta + 31.35 = 0$

This equation can be solved for $\tan \theta$ using the quadratic formula to give:

$\tan \theta = 4.7665$ and $\tan \theta = 0.33557$

From this we can see that the angle needed is:

$\theta = 78.15$ and $\theta = 18.55$

So the answer is:

$\theta = 18.55$

Assuming there is no air resistance, the movement along horizontal axis is uniform, while along vertical axis it is uniformly accelerated.

Writing down the equations for both the motions one gets

$\begin{cases}x=vt\cos{\vartheta} \\ y=vt\sin{\vartheta}-\frac{g}{2}t^2 \end{cases}$

Substituting generic $x$ , $y$ and $t$ for those at the moment of a hit and solving simultaneous equations for $\vartheta$ excluding $T$ one gets $Y = X\tan{\vartheta} - \frac{g X^2}{2 v^2}\frac{1}{\cos^2{\vartheta}}$ or, reducing all the trigonometric functions to $\tan$ : $Y = X\tan{\vartheta} - \frac{g X^2}{2 v^2}(1+\tan^2{\vartheta})$ .

Solving this quadratic equation for $\tan{\vartheta}$ , one gets $\tan{\vartheta} = \frac{v^2}{gX} \pm \sqrt{\frac{v^4}{g^2 X^2}-1-\frac{2 Y v^2}{g X^2}}$ , and $\vartheta\in\{18.55, 78.15\}^\circ$ .

For a projectile whose given parameters are the initial velocity $v_{o}$ , vertical and horizontal displacement $x$ and $y$ , the appropriate solution would be

$y = x\tan \theta - \frac {gx^{2}}{2v_{o}^{2} \cos^{2} \theta}$

Substituting the given values $x = 100$ $m$ , $y = 11.75$ $m$ and $v_{o} = 50$ $m/s$ , we get $\theta = 18.55^ \circ$ .