Well, well, well

Let me post a different solution than posted by @Sadi Kneipp Neto . The whole point is that the particle should 'come out' of the well. That means it is not sufficient to only travel a vertical distance $h = 15 m$ but it should also travel a horizontal distance $r = 8 m$ . Our old friend 'energy conservation' helps us to calculate the minimum value of the vertical component of velocity required to traverse distance $h$ . Equating potential energy at the top to the kinetic energy at the bottom, we get, $mgh = \frac{1}{2}mv_{y}^{2}$ This gives, $v_{y}=\sqrt{2gh}$ The amount of time $t$ it takes to traverse a distance $h$ with this initial speed is then given by: $0 = v_{y}-gt$ which gives $t=\sqrt{\frac{2h}{g}}$ During exactly this much time, particle must travel a distance $r$ to reach out of well. Hence the horizontal component of velocity that is required is: $v_{x}=\frac{r}{\sqrt{\frac{2h}{g}}} = r\sqrt{\frac{g}{2h}}$ Hence the magnitude of the velocity that is required is: $v = \sqrt{\frac{r^{2}g}{2h} + 2gh} = \boxed{17.9 m/s}$

$17.89 \text{ ms}^{-1}$ initial launch speed is really easy for me, so I managed to get out and open my computer to write this solution.

From the problem, we get two equations:

$8 = u\cos\theta t$

$15 = u\sin \theta t - 5t^2$

From the first equation, we get $t = \frac{8}{u\cos \theta}$ and substitute into the second equation, we get

$\displaystyle 15 = 8 \tan \theta - \frac{320}{u^2 \cos ^2 \theta}$

$\displaystyle u = \sqrt{\frac{320\sec ^2 \theta}{8\tan \theta - 15}}$

The next step is a simple but tedious derivation. You'll get to the answer $u = 17.89 \text{ms}^{-1}$ .

The solution becomes quite simple once the concept of Safety Parabola is known.

Setting the axis to the launch point, our parabola needs to pass through the well's top edge.

Parabola: $y=\frac{-gx²}{2V²} + \frac{V²}{2g}$

Plugging x=8 and y=15, we arrive at a biquadratic equation.

${ V }^{ 4 }-300{ V }^{ 2 }-6400=0$

Solving for V, we get V=17.89

The positive displacement upwards can be calculated as follows:

$d_{y} = \frac{at^2}{2} + v_{y} = 15$

so we get:

$v_{y} = \frac{15+5t^2}{t}$

For the velocity in the x direction we get:

$v_{x} = 8/t$

The velocity can be expressed in terms of t:

$v= \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(8/t)^2 + (\frac{15+5t^2}{t})^2}$

Simplifying, we get:

$v= \sqrt{289/t^2 + 25t^2+150}$

Since we're looking for a minimum velocity, we can differentiate v with respect to t to find the t at which v is minimal.

$\frac{dv}{dt} = 0 = \frac{-2 \frac{289}{t^3} +50t}{2\sqrt{289/t^2 + 25t^2+150}}$ => t = 1.8439 s

Plugging t back into the equation for v will give you the velocity 17.89

Assuming that Christopher JUST makes it out of the well (i.e. the maximum height achieved by him during his jump is $15$ m right at the edge of the well), we can say that half of the horizontal range of his jump is equal to the radius of the well(by symmetrey).

So, using what we know from projectile motion we can say that

$\frac {1}{2} \cdot \frac {v^{2}\sin (2\theta)}{g} = 8, \frac {v^{2}\sin^{2}\theta}{2g} = 15$

where $\theta$ is his launch angle with respect to the horizontal and $v$ is his velocity at launch.

$\therefore \frac {\sin(2\theta)}{\sin^{2}\theta} = 2\cot \theta = \frac {8}{15} \Rightarrow \theta = \cot^{-1} \left(\frac {8}{30}\right) = 75.0685828219^{\circ}$

So we can conclude that

$v = \frac {\sqrt {300}}{\sin \theta} \approx \boxed {17.92 \text {m}\text {s}^{-1}}$