Correct option is (D)

From energy conservation $\to$

$\begin{array}{rl}& mg(R+R\sin 45)=\frac{1}{2}m{v}^2\\ & \Rightarrow 10(1+\frac{1}{\sqrt{2}})\times 14=\frac{1}{2}{v}^2\\ & \Rightarrow 10(1+\frac{\sqrt{2}}{2})\times 28=v^2\\ & \Rightarrow 10(1+0.7)\times 28=v^2\\ & \Rightarrow v=21.81\end{array}$

Correct option is (A)

To solve this problem, we need to analyze both the energy loss and the initial velocity of the rubber ball before it strikes the ground.

First, let's consider the energy loss. The energy involved here is gravitational potential energy. The initial potential energy of the ball when it is about to fall is given by $U_i=mgh$, where $U_i$ is the initial potential energy, $m$ is the mass of the ball, $g$ is the acceleration due to gravity, and $h$ is the initial height from which the ball falls. After the ball rebounds, it reaches a height of $h/2$. The potential energy at this new height is $U_f=mg\cdot \frac{h}{2}$.

The energy loss can be calculated as the difference between the initial and final potential energies, and to find the percentage energy loss, we divide this difference by the initial energy and multiply by 100:

$\text{Energy loss percentage}=\frac{(U_i-U_f)}{U_i}\times 100$

Substituting the values of $U_i$ and $U_f$ gives:

$\text{Energy loss percentage}=\frac{(mgh-mg\frac{h}{2})}{mgh}\times 100$

By simplifying, we find:

$\text{Energy loss percentage}=\frac{mgh-\frac{1}{2}mgh}{mgh}\times 100=\frac{1}{2}\times 100=50\mathrm{\%}$

This tells us that the energy loss percentage is indeed $50\mathrm{\%}$.

Next, we'll find the velocity of the ball just before it strikes the ground. The velocity can be determined using the formula for the velocity of an object in free fall:

$v=\sqrt{2gh}$

Here, $v$ is the velocity of the ball just before impact, $g$ is the acceleration due to gravity, and $h$ is the height from which the ball falls. This formula shows that the initial velocity of the ball before it strikes the ground is $\sqrt{2gh}$, not taking into account air resistance and assuming it starts from rest.

Therefore, the correct answer is Option A: $50\mathrm{\%}$, $\sqrt{2gh}$.

Correct option is (C)

By COME

$\begin{array}{rl}& {\mathrm{KE}}_{\mathrm{A}}+{\mathrm{U}}_{\mathrm{A}}={\mathrm{KE}}_{\mathrm{B}}+{\mathrm{U}}_{\mathrm{B}}\\ & 0+\mathrm{mg}(1)=\frac{1}{2}{\mathrm{mv}}^2+\mathrm{mg}\times 0.5\\ & v=\sqrt{g}=\sqrt{10}\mathrm{m}/\mathrm{s}\end{array}$

Correct option is (A)

Apply energy conservation between A & B

$\begin{array}{rl}& \frac{1}{2}{\mathrm{mV}}_{\mathrm{L}}^2=\frac{1}{2}{\mathrm{mV}}_{\mathrm{H}}^2+\mathrm{mg}(2\mathrm{L})\\ & \because {\mathrm{V}}_{\mathrm{L}}=\sqrt{5\mathrm{gL}}\end{array}$

So, ${\mathrm{V}}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$

$\frac{(\mathrm{K}.\mathrm{E}{)}_{\mathrm{A}}}{(\mathrm{K}.\mathrm{E}{)}_{\mathrm{B}}}=\frac{\frac{1}{2}\mathrm{m}(\sqrt{5\mathrm{gL}}{)}^2}{\frac{1}{2}\mathrm{m}(\sqrt{\mathrm{gL}}{)}^2}=\frac{5}{1}$

Correct option is (B)

$\begin{array}{rl}& v^2-u^2=2aS\\ & {\left(\frac{2u}{3}\right)}^2=u^2+2(-a)(4\times {10}^{-2})\\ & \frac{4u^2}{9}=u^2-2a(4\times {10}^{-2})\\ & -\frac{5u^2}{9}=-2a(4\times {10}^{-2})\dots (1)\\ & 0={\left(\frac{2u}{3}\right)}^2+2(-a)(x)\\ & -\frac{4u^2}{9}=-2ax\dots (2)\end{array}$

$(1)/(2)$

$\begin{array}{rl}& \frac{5}{4}=\frac{4\times {10}^{-2}}{\mathrm{x}}\\ & \mathrm{x}=\frac{16}{5}\times {10}^{-2}\\ & \mathrm{x}=3\cdot 2\times {10}^{-2}\mathrm{m}\\ & \mathrm{x}=32\times {10}^{-3}\mathrm{m}\end{array}$