The Midair Grab

Let’s place the origin of the Cartesian axes on the brink of left ledge. Now, the system Artefact-Indiana Jones, at time $t=0$ , will be at $P(X_{cm},0)$ :

1) $X_{cm}=\frac {m \times 0 + 2m\times 2D}{3m}=\frac{4}{3}D$

and its momentum on the X-axis is

2) $3mv_{cmx}=mv_{0A}-2mv_{0I}$

from which we get

3) $v_{cmx}=\frac{v_{0A}-2v_{0I}}{3}$

Since the position we have to reach is D the horizontal distance to be covered is, by 1)

4) $D-\frac{4}{3}D =-\frac{1}{3}D$

namely:

5) $v_{cmx}\times t=-\frac{1}{3}D$

Then, remembering that the time an object takes to fall is independent from its mass and is function uniquely of height, we can write

6) $H= -\frac{1}{2}gt^2$

So, from 6) and 5) we get

7) $\frac{v_{0A}-2v_{0I}}{3}\sqrt{\frac{2H}{g}}=-\frac{1}{3}D$

and, at last

$\boxed{ v_{0I}=\frac{1}{2}(D \sqrt{\frac{g}{2H}}+v_{0A})}$

Let's place the origin of our (x,y) coordinate system at the point directly below the artifact's ledge, and directly to the left of the platform. Then, the artifact's x- and y- position vector, as a function of time, can be given as:

$\vec{r}_{A} (t) = <(v_{0A}t) , (H-\frac{1}{2}gt^{2})>$

Its velocity vector, as a function of time, can be given as:

$\vec{v}_{A} (t) = <(v_{0A}) , (-gt)>$

Similarly, Indy's position vector function can be written:

$\vec{r}_{I}(t) = <(2D - v_{0I}t) , (H - \frac{1}{2}gt^{2})>$

His velocity vector function can be written:

$\vec{v}_{I}(t) = <(-v_{0I}) , (-gt)>$

Now, because Indy left the ledge at the same instant the artifact was thrown, and they both had initially horizontal trajectories, he is garunteed to catch the artifact, no matter his starting speed (provided it doesn't hit the lava first).

Let's figure out when and where the grab will take place. We set the x- and y- components of Indy's and the artifact's position vectors equal to eachother. Doing this gives us the time and position of the grab as:

$t_{G} = \frac{2D}{v_{0A}+v_{0I}}$ , and $\vec{r}_{G} = <(\frac{2v_{0A}D}{v_{0A}+v_{0I}}) , (H - \frac{2gD^{2}}{(v_{0A}+v_{0I})^{2}})>$

Now, when the grab takes place, Indy's momentum and the artifact's momentum are going to combine. We perform a vector addition of their momentums at the time of the grab:

$3m\vec{v}_{G} = <(mv_{0A}-2mv_{0I}) , (m(\frac{-2gD}{v_{0A}+v_{0I}})+2m(\frac{-2gD}{v_{0A}+v_{0I}}))>$

Simplifying we get:

$\vec{v}_{G} = <(\frac{v_{0A}-2v_{0I}}{3}) , (\frac{-2gD}{v_{0A}+v_{0I}})>$

Now we have equations representing the position $\vec{r}_{G}$ and velocity $\vec{v}_{G}$ at the time of the grab. Let's "reset the clock," so that $t_{G}$ corresponds to $t = 0$ .

We know that Indy and the artifact need to end up at the position $\vec{r}_{p} = <D,0>$ , the position of the platform. We create a new parametric vector equation for their position as a function of time:

$\vec{r}(t) = <(r_{Gx}+v_{Gx}t),(r_{Gy}+v_{Gy}t-\frac{1}{2}gt^{2})>$

By setting the y- component of this vector equal to 0, and then using the quadratic formula, we find that Indy and the artifact land on the platform at time:

$t_{p} = \frac{v_{Gy}+\sqrt{v_{Gy}^{2}+2gr_{Gy}}}{g}$

We set the x-component equal to $D$ , and plug in this formula for $t_{p}$ , to get:

$D = r_{Gx} + v_{Gx}[\frac{v_{Gy}+\sqrt{v_{Gy}^{2}+2gr_{Gy}}}{g}]$

We make the following substitutions:

$r_{Gx} = \frac{2v_{0A}D}{v_{0A}+v_{0I}}$

$r_{Gy} = H - \frac{2gD^{2}}{(v_{0A}+v_{0I})^{2}}$

$v_{Gx} = \frac{v_{0A}-2v_{0I}}{3}$

$v_{Gy} = -\frac{2gD}{v_{0A}+v_{0I}}$

From there, we solve the equation for $v_{0I}$ (it's quite a bit of work, but it's doable):

$v_{0I} = \frac{v_{0A}}{2}+\frac{D}{2}\sqrt{\frac{g}{2H}}$

Plugging in the numbers, we get:

$v_{0I} = \frac{20.2 \frac{m}{s}}{2} + \frac{15.3 m}{2}\sqrt{\frac{9.81 \frac{m}{s^{2}}}{2(10.1 m)}} = \boxed{v_{0I} = 15.4 \frac{m}{s}}$