Reach for the Summit - P-S1-A3

Since both balls are thrown horizontally. they drop to the ground at the same time. And since ball $2$ also reaches point $B$ after a bounce. This means that ball $2$ takes $3$ times the time taken by ball $1$ to reach point $B$ or the horizontal velocity of ball $2$ is $\frac 13$ that of ball $1$ .

Now if ball $1$ takes time $t$ to drop from height $H$ to $h$ , then iball $2$ will take $3t$ to drop from $H$ to the ground and rebound to height $h$ . Then for ball $1$ to drop from $H$ to $h$ , we have $H-h = \frac 12 gt^2$ . Ball $2$ , will takes $2t$ to reach the ground, therefore the velocity upon reaching the ground is $v = 2gt$ . Ball $2$ takes $t$ to reach $h$ , therefore $h = vt - \frac 12 gt^2 = 2gt^2 - \frac 12 gt^2 = \frac 32 gt^2$ .

Now we have: $H - h = \frac 12 gt^2 = \frac 13 h \implies H = \frac 43 h \implies h = \frac 34 H$ . Therefore $\lambda = 0.75$ and $\lfloor 1000\lambda \rfloor = \boxed{750}$ .

By the principle of reversibility of path $A$ is the mid point of $\overline {CB}$ , where $C$ is the point vertically below the point $O$ on the ground.

Then $|\overline {CA}|=\dfrac {3}{2}|\overline {CD}|$ , where $D$ is the point where the second ball bounces from.

We have the following equations to deal with :

$h=\dfrac {3ga^2}{2u^2}$ , where $a=|\overline {CA}|,u=$ velocity of projection of the first ball.

$u=2a\sqrt {\dfrac {g}{2H}}$

$h=\dfrac {a}{u}\sqrt {2gH}-\dfrac {ga^2}{2u^2}$

Eliminating $a$ and $u$ from these we get

$h=H\times \dfrac {1-2-4+8}{4+1-2+5-4}=\dfrac {3H}{4}$

Hence the required answer is $\boxed {750}$ .

Aliter:

Let $C$ be the point on ground vertically below $O$ , $D$ be the point where the second ball bounces for the first time, $|\overline {CD}|=c$

Then $|\overline {CB}|=3c$

Equation of path of the first ball is

$y=H-\dfrac {Hx^2}{9c^2}$

Equation of path of the second ball after bouncing from $D$ is

$y=3(x-c)-\dfrac {H}{c^2}(x-c) ^2$

So we have $\dfrac {3c^2}{H}+c=3c$

$\implies H=\dfrac {3c}{2}$

Let the two paths intersect at $(p, h)$

Then

$h=H-\dfrac {Hp^2}{9c^2}=3(p-c)-\dfrac {H}{c^2}(p-c) ^2$

Substituting for $H$ and solving for $p$ we get

$p=3c,\dfrac {3c}{2}$

The first value corresponds to the point $B$ where $h=0$

The second value yields

$h=H-\dfrac {H}{4}=\dfrac {3H}{4}$

So, $\dfrac {h}{H}=\dfrac {3}{4}=0.75$

Hence the required answer is $\boxed {750}$ .