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

The correct option is 5√(√2 + 1) m/s.

1. Understanding the step coordinates:
From the given diagram:

Each step has a horizontal width of $0.5\text{m}$ and a vertical height of $0.5\text{m}$. Taking the starting point $O$ as the origin $(0,0)$, the coordinates $(x,y)$ of the corner of the 5th step are:

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

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

2. Deriving the minimum projection speed:
The trajectory equation of a projectile launched with speed $u$ at an angle $\theta$ is:

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

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

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

For a real angle of projection $\theta$ to exist to reach the target point $(x,y)$, the discriminant of this quadratic equation must be greater than or equal to zero:

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

Simplifying this inequality yields:

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

Solving for the minimum value of $u^2$:

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

3. Calculating the numerical value:
Substituting $x=2.5\text{m}$, $y=2.5\text{m}$, and acceleration due to gravity $g=10{\text{m/s}}^2$:

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

$u_{\text{min}}=\sqrt{10\left(2.5+2.5\sqrt{2}\right)}$

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

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