projectors

Relevant wiki: Projectile Motion

Let the distance between the point of projection and the wall be $d_1$ and that between the wall and the crown be $d_2$ .

Now, $d_1$ is the distance between point of projection and the highest point in the path of projectile. So we have ( $T$ is the time of flight of the complete projectile motion of the ball),

$d_1 = v_1 \cos \theta \cdot \dfrac T2 = \left( v_1 \cos \theta \right) \times \left( \dfrac{1}{2} \cdot \dfrac{2v_1 \sin \theta}{g} \right) = \dfrac{v_1^2 \sin 2\theta}{2g}$

The altitude of the highest point (where it hits the wall) is

$h = v_1 \sin \theta \cdot \dfrac T2 - \dfrac 12 g {\left( \dfrac T2 \right)}^2 \implies h = v_1 \sin \theta \left( \dfrac{v_1 \sin \theta}{g} \right) - \dfrac 12 g \cdot {\left( \dfrac{v_1 \sin \theta}{g} \right)}^2 \implies h = \dfrac{v_1^2 \sin^2 \theta}{2g}$

After the projectile hits the wall (at exactly normal to the surface), the vertical velocity of the ball becomes zero and so the time it takes to come down is

$h = \dfrac 12 g t^2 \implies t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2}{g} \cdot \dfrac{v_1^2 \sin^2 \theta}{2g}} = \dfrac{v_1 \sin \theta}{g}$

Thus

$d_2 = \dfrac{v_1}{2} \cdot t = \dfrac{v_1^2 \sin \theta}{2g}$

The required length $\ell$ is expressed as

$\ell = d_1 - d_2 = \dfrac{v_1^2}{2g} \left( \sin 2 \theta - \sin \theta \right) = \dfrac{20^2}{2 \times 9.81} \left( \sin \dfrac{\pi}{3} - \sin \dfrac{\pi}{6} \right) \approx \boxed{7.4623}$

Relevant wiki: Projectile Motion

At the wall, the ball's horizontal speed changes from $10\sqrt{3}\ \text{m/s}$ to $10\ \text{m/s}$ .

Therefore instead of returning to its launch point, the ball only travels back $\frac{1}{\sqrt{3}}$ times as far.

The range of the projectile is $\frac{2v^2\sin\theta \cos\theta}{g}\approx35.3\text{m}$

So $l=\frac{35.3}{2}(1-\frac{1}{\sqrt{3}})\approx\boxed{7.46\text{m}}$ .

We will work with Aziz's diagram, adding d, the distance from the launch point to the wall. Notice that the time to teach the top of the trajectory and the time to fall back down from the wall to the crown will be the same. Let's call it T.

First let's apply the equation for uniform acceleration " $v=at$ " to the vertical component of motion, starting at the launch point and finishing at the top of the trajectory. This gives the equation

$V \sin{\theta}=gT$

Now let's apply the equation for unaccelerated motion " $s=ut$ " to the horizontal component of the motion. First of all from the launch point to the wall, giving

$d=VT \cos{\theta}$

Then from the wall to the crown giving

$d-L=\frac{VT}{2}$

Simple algebra with these three equations allows us to eliminate d and T to get

$L=\frac{V^{2} \sin{\theta}}{g}\left(\cos{\theta}-\frac{1}{2}\right)$

Finally put in the numbers to get

$L=\frac{20^{2}\sin{\frac{\pi}{6}}}{9.81}\left(\cos{\frac{\pi}{6}}-\frac{1}{2}\right)=\boxed{7.462}$

So, what we have and what we need.

We have:

• $v_1 = \SI[per-mode=symbol]{20}{\meter\per\second}$
• $\theta = \SI{\frac{\pi}6}{\radian}$
• $g = \SI[per-mode=symbol]{9.81}{\meter\per\second\squared}$
• At the peak of its trajectory, the ball hits the wall and reflects with speed $\frac12 v_1.$

We need:

• $l$

We can define $l$ as $l = A - B$ .

Thus we have to gather information about two distances: $A$ and $B$ .

Let's find A firstly. In the horizontal direction, the motion is described by the uniform velocity motion (UVM) equations of motion (acceleration in horizontal direction is $0$ ):

• $x(t) = v_{ix}t + x_i$
• $v_x(t) = v_{ix}$

$x_i$ is an initial position and $v_{ix}$ is an initial velocity.

In the vertical direction, the motion is described by the uniformly accelerated motion (UAM) equations of motion (there is an acceleration ( $g = \SI[per-mode=symbol]{9.81}{\meter\per\second\squared}$ )):

• $y(t) = \frac{1}{2}(-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t^2 + v_{iy}t + y_i$
• $v_y(t) = (-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t + v_{iy}$

$y_i$ is an initial position and $v_{iy}$ is an initial velocity.

We know that the ball hit the wall at the peak of its trajectory (when the ball reaches its peak, it will have zero velocity in the $y$ -direction), so it was when $v_y(t) = 0$ .

• \(v_{ix} = v_1\cos \theta = \SI[per-mode=symbol]{20\cos (\SI{\frac{\pi}6})}{\meter\per\second} = \SI[per-mode=symbol]{10\sqrt{3}}{\meter\per\second}\)
• \(v_{iy} = v_1\sin \theta = \SI[per-mode=symbol]{20\sin (\SI{\frac{\pi}6})}{\meter\per\second} = \SI[per-mode=symbol]{10}{\meter\per\second}\)

Knowing these guys we can go further.

• $v_y(t) = (-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t + v_{iy}$
• $0 = (-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t + v_{iy}$
• $0 = (-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t + \SI[per-mode=symbol]{10}{\meter\per\second}$
• $(-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})t = - \SI[per-mode=symbol]{10}{\meter\per\second}$
• \(t = \SI[per-mode=symbol]{\frac{10}{\SI[per-mode=symbol]{9.81}{\meter\per\second\squared}}}{\meter\per\second}\)
• $t = \SI[per-mode=symbol]{1.019}{\second}$

Now we know we hit the wall after $\SI[per-mode=symbol]{1.019}{\second}$ . So, the distance is:

• $x(\SI[per-mode=symbol]{1.019}{\second}) = (\SI[per-mode=symbol]{10\sqrt{3}}{\meter\per\second})(\SI[per-mode=symbol]{1.019}{\second}) + 0 = \SI[per-mode=symbol]{17.649}{\meter}$
• $A = \SI[per-mode=symbol]{17.649}{\meter}$

And the height:

• $h_{max} = y(\SI[per-mode=symbol]{1.019}{\second}) = \frac{1}{2}(-\SI[per-mode=symbol]{9.81}{\meter\per\second\squared})(\SI[per-mode=symbol]{1.019}{\second})^2 + (\SI[per-mode=symbol]{10}{\meter\per\second})(\SI[per-mode=symbol]{1.019}{\second}) + 0$
• $h_{max} = 5.0968$

Let's find B secondly. The ball hit the wall. What is going on now? The velocity has changed. Now we have:

• $v_1 = \frac{\SI[per-mode=symbol]{20}{\meter\per\second}}{2} = \SI[per-mode=symbol]{10}{\meter\per\second}$

When the ball hits the crown? When $y(t) = 0$ (the ball will hit the ground in this time). So, we have to calculate $0 = \frac{1}{2}(-9.81)t^2 + v_{iy}t + y_i$ . We need information about $y_i$ and $v_{iy}$ .

• \(v_{iy} = v_1\sin 0 = \SI[per-mode=symbol]{10\sin (\SI{0})}{\meter\per\second} = 0\)
• $y_i = h_{max} = 5.0968$

Now, we can find a solution for this guy:

• $y(t) = \frac{1}{2}(-9.81)t^2 + v_{iy}t + y_i$
• $0 = \frac{1}{2}(-9.81)t^2 + 0t + 5.0968$
• $0 = \frac{1}{2}(-9.81)t^2 + 5.0968$
• $t = \SI[per-mode=symbol]{1.01936}{\second}$

So, we know that the ball will hit the crown after $\SI[per-mode=symbol]{1.01936}{\second}$ (after hitting the wall). Let's put this into $x(t) = v_{ix}t + x_i$ . But, before we need to know the value of $v_{ix}$ .

• \(v_{ix} = v_1\cos \theta = \SI[per-mode=symbol]{10\cos (\SI{0})}{\meter\per\second} = \SI[per-mode=symbol]{10}{\meter\per\second}\)

Now we can figure out the value of $x(\SI[per-mode=symbol]{1.01936}{\second})$ :

• $B = x(\SI[per-mode=symbol]{1.01936}{\second}) = (\SI[per-mode=symbol]{10}{\meter\per\second})(\SI[per-mode=symbol]{1.01936}{\second}) + 0$
• $B = x(\SI[per-mode=symbol]{1.01936}{\second}) = \SI[per-mode=symbol]{10.1936}{\meter}$

We have $A$ and $B$ , so we can find out $l = A - B$ :

• $l = \SI[per-mode=symbol]{17.649}{\meter} - \SI[per-mode=symbol]{10.1936}{\meter} = \SI[per-mode=symbol]{7.46}{\meter}$