Stones, Up and Down

When the stone was $h$ above A, we can write: $v^2=u^2-2gh \implies 4v^2=4u^2-8gh \dots (1)$ .When the stone was $h$ below A we can write $4v^2=u^2+2gh \dots (2)$ . Eliminating $4v^2$ from both the equations we have: $3u^2-10gh=0 \implies u^2=\dfrac{10gh}{3}$ . We also know $h_{max}=\dfrac{u^2}{2g}=\dfrac{10gh}{6g}=\dfrac{5h}{3}$ giving $p+q=5+3=\boxed{8}$ .

Since the kinetic energy of a projectile is $T = \frac{1}{2}mv^{2}$ , we know that the stone has four times as much kinetic energy at a distance $h$ below $A$ as it does at a distance $h$ above $A$ . These two points have a height difference of $2h$ , so an energy balance gives us $T_{high} + 2mgh = T_{low} = 4T_{high}$ , or $T_{high} = \frac{2}{3}mgh$ . Since the stone reaches its greatest height when all its kinetic energy is converted to gravitational potential energy, this point must be $\frac{2}{3}h$ higher than the point $h$ above $A$ , for a total height of $\frac{5h}{3}$ . Thus the answer is $p + q = 5 + 3 = \boxed{8}$ .

Let the initial upward speed of the stone be $u$ . Then the height of the stone relative to $A$ at time $t$ is

$y = ut - \dfrac{g}{2}t^{2} \Longrightarrow \dfrac{g}{2}t^{2} - ut + y = 0 \Longrightarrow t = \dfrac{u \pm \sqrt{u^{2} - 2gy}}{g}$ , (i).

Now the velocity of the stone at time $t$ is $v(t) = u - gt$ . Plugging in equation (i) gives us

$v(y) = u - (u \pm \sqrt{u^{2} - 2gy}) = \pm \sqrt{u^{2} - 2gy}$ .

Now for $h \gt 0$ we are given that

$|v(-h)| = 2|v(h)| \Longrightarrow \sqrt{u^{2} + 2gh} = 2\sqrt{u^{2} - 2gh}$

$\Longrightarrow u^{2} + 2gh = 4(u^{2} - 2gh) \Longrightarrow 10gh = 3u^{2} \Longrightarrow h = \dfrac{3u^{2}}{10g}$ .

Now the maximum height of the stone is achieved when

$v(t) = u - gt = 0 \Longrightarrow t = \dfrac{u}{g} \Longrightarrow y_{max} = \dfrac{u^{2}}{g} - \dfrac{g}{2}*\dfrac{u^{2}}{g^{2}} = \dfrac{u^{2}}{2g}$ .

So $\dfrac{y_{max}}{h} = \dfrac{\dfrac{u^{2}}{2g}}{\dfrac{3u^{2}}{10g}} = \dfrac{5}{3}$ , and thus $p + q = 5 + 3 = \boxed{8}$ .

Let $H$ be the highest point, and $P$ and $Q$ the two points discussed in the problem.

Let $t$ be the time it takes from $P$ to $Q$ , and let $v$ the speed at point $P$ . Then the speed at point $Q$ is $2v$ .

Since speed is directly proportional to falling time,

• the falling time between $H$ and $P$ is also $t$ ;

• the average speed between $P$ and $Q$ is $1\tfrac 12 v$ ;

• the average speed between $H$ and $P$ is $\tfrac 12 v$ .

Thus the ratio of distances is $HP : PQ = \tfrac12 : 1\tfrac12 = 1:3$ . It follows that $HP = \tfrac13\cdot 2h = \tfrac 23 h$ , so that $H$ lies at $5h/3$ above $A$ .

Let the velocities at height $h$ above $A$ and height $h$ below $A$ be $v_1$ and $2v_1$ respectively. If the object is thrown upwards with a velocity $v_0$ , then

$v_1^{2} = v_0^{2} - 2gh$

and

$(2v_1)^{2} = v_0^{2} + 2gh \Rightarrow v_1^{2} =\frac {4gh}{3}$

Therefore,

$v_0^{2} = \frac {10gh}{3} \therefore H_{max} = \frac {v_0^{2}}{2g} = \frac {5h}{3}$