Consider a series of steps as shown. A ball is thrown from O. Find the minimum speed of directly jump to 5th step.

1. Identify the Coordinates of the Target Point
From the given image, each step has a width of 0.5 m and a height of 0.5 m.
To directly jump to the 5th step from the origin O(0,0), the ball must reach the corner of the 5th step.
The coordinates of this corner point $(x,y)$ are:

$x=5\times 0.5\text{ m}=2.5\text{ m}$

$y=5\times 0.5\text{ m}=2.5\text{ m}$

2. Derivation of Minimum Speed to Reach a Point (x, y)
The equation of trajectory for a projectile launched with speed u at an angle $\theta$ to the horizontal is:

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

Using the trigonometric identity $\frac{1}{{\cos }^2\theta }=1+{\tan }^2\theta$, we rewrite this as a quadratic equation in terms of $\tan \theta$:

$y=x\tan \theta -\frac{g{x}^2}{2u^2}(1+{\tan }^2\theta )$

Rearranging the terms:

$\frac{g{x}^2}{2u^2}{\tan }^2\theta -x\tan \theta +(y+\frac{g{x}^2}{2u^2})=0$

For a real projection angle $\theta$ to exist, the discriminant of this quadratic equation must be non-negative ($D\ge 0$):

$x^2-4\left(\frac{g{x}^2}{2u^2}\right)(y+\frac{g{x}^2}{2u^2})\ge 0$

Dividing by $x^2$ and simplifying:

$1-\frac{2g}{u^2}(y+\frac{g{x}^2}{2u^2})\ge 0$

$u^4-2gy{u}^2-g^2{x}^2\ge 0$

Solving this inequality for $u^2$ yields the condition for the minimum required speed:

$u_{\text{min}}^2=g(y+\sqrt{x^2+y^2})$

3. Calculation
Substitute the values $x=2.5\text{ m}$, $y=2.5\text{ m}$, and acceleration due to gravity $g=10\text{ m/s}{g}^2$:

$u_{\text{min}}^2=10\times (2.5+\sqrt{{2.5}^2+{2.5}^2})$

$u_{\text{min}}^2=10\times (2.5+2.5\sqrt{2})$

$u_{\text{min}}^2=25(\sqrt{2}+1)$

Taking the square root to find the minimum launch speed:

$u_{\text{min}}=5\sqrt{\sqrt{2}+1}\text{ m/s}$