Bird Problem

since all parabolas are similar, the exact number isnt relevant, just the ratio. consider $y=-x^2$ , the problem is reduce to the ratio $a:b$ , which is $\frac{2}{1+\sqrt2}$ .

edit: thanks to Meredith Hughes for pointing out the y-coordinate should be -2

The ball follows a path of a parabola that opens downward, which has a general form of $y = -ax^2 + bx + c$ . If we let the maximum height of the ball be the origin, then $b = 0$ and $c = 0$ and the general form simplifies to $y = -ax^2$ .

Laurie runs on a $y$ -value of $-h$ , which intersects the parabola when $-h = -ax^2$ , or $x = \pm k$ for $k = \sqrt{\frac{h}{a}}$ . The distance she runs is therefore $d_L = 2k$ , at a speed of $s_L = \frac{2k}{t}$ over time $t$ .

Orlando is standing on a $y$ -value of $-2h$ , which intersects the parabola when $-2h = -ax^2$ , or $x = \pm k\sqrt{2}$ for $k = \sqrt{\frac{h}{a}}$ . The horizontal distance of the ball is therefore $d_b = k + k\sqrt{2}$ , at a speed of $s_b = \frac{k + k\sqrt{2}}{t}$ over time $t$ .

The ratio of Laurie's speed to the ball's horizontal speed is $\frac{s_L}{s_b} = \frac{\frac{2k}{t}}{\frac{k + k\sqrt{2}}{t}}$ $=$ $\frac{2}{1 + \sqrt{2}}$ $=$ $2\sqrt{2} - 2 \approx \boxed{0.8284}$ .

• First consider the ball's vertical velocity, $v_{y}=\sqrt{2\cdot g\cdot 2h}=2\sqrt{gh}$
• Now consider the 2 times when the ball's vertical displacement is $h$ .
• $h=2\sqrt{gh}\cdot t-\frac{1}{2}gt^{2}$
• This gives, $t=(2\pm \sqrt{2})\sqrt{\frac{h}{g}}$
• Now let the ball's horizontal velocity be $v_{b}$ , so the distance moved by Laurie will be the difference of the horizontal distances moved by the ball during the 2 times, i.e, $d=v_{b}\cdot 2\sqrt{\frac{2h}{g}}$
• This is also the distance moved by Laurie during the entire time, thus $d=v_{L}\cdot (2+\sqrt{2})\sqrt{\frac{h}{g}}$
• Equating the 2 equations, $\frac{v_{L}}{v_{b}}=\frac{2\sqrt{2}}{2+\sqrt{2}}=\frac{2}{1+\sqrt{2}}=0.8284271247..$

The ball travels in a parabolic path. The distance from the turning point to Laurie is $\sqrt h$ , and the distance to Orlando is $\sqrt {2h}$ . Therefore the ball travels $(1+\sqrt 2)\sqrt h$ and Laurie travels $2\sqrt h$ . The ratio of Laurie's speed to the ball's speed is the ratio of the distances $=\frac{2}{1+\sqrt2}\approx 0.8284$ .

First, we can see that the time that the ball spent to cover the second and the third part of the curve (over the cliff) is the same $t$ . Now, we can say that the speed at the end of the trajectory is the same as at the symmetrical point. $V=0+gt=gt$ . The time that the ball spent to climb the cliff is $t_2$ which could be obtained from:

$\\h=0*t+\frac{gt_2}{2}$

$\\h=Vt_2+\frac{gt^2_2}{2}=gtt_2+\frac{gt^2_2}{2}$

From that two equations we can find that

$\\\frac{gt_2}{2}=gtt_2+\frac{gt^2_2}{2} -> t_2=(\sqrt{2}-1)t$

Therefore, the ball spent $2t$ to cover the distance $s$ that Laurie covers by $2t+t_2=(1+\sqrt(2))t$ . Hence, we can find the relation

$\\s=v_{ball}*2t$

$\\s=v_{Laurie}*(\sqrt{2}+1)t$

Finally, we divide the second one by first:

$\\1=\frac{v_{Laurie}*(\sqrt{2}+1)t}{v_{ball}*2t}=\frac{v_{Laurie}*(\sqrt{2}+1)}{v_{ball}*2}$

$\frac{v_{Laurie}}{v_{ball}}=\frac{2}{\sqrt{2}+1}$

It is equal to ratio of (t1-t2)/t1, where t1 and t2 are two times to reach half the height. The equation to be solved for t1 and t2 is the quadratic equation for achieving half height (h/2) when the ball is thrown upward under gravity.

(1/2)g(t^2) - √(2gh)(t) + (h/2)=0

{(t1-t2)/t1} = {2/(1+√2)} = 0.8284