Kinematics-1D

$s_1 = \frac{1}{2} \times 5 \times (5)^2 = \frac{125}{2} m$
$\mathrm{v} = 0 + 5 (5) = 2 5 \mathrm{m/s}$
$\mathrm{s} _ {2} = 2 5 \times 1 0 = 2 5 0 \mathrm{m}$
$\begin{array}{c c c c c} u = 0 & & v \\ \text {A} & s _ {1} & B & s _ {2} & C \\ t = 0 & & t = 5 \mathrm{sec} & & t = 1 5 \mathrm{sec} \end{array}$
Total distance travelled = $s_{1} + s_{2} = 312.5$ m

$a=\frac{20-10}{5}=2m/s^{2}$ for 2 seconds before $t_{0}$ $10 = v + 2(2)$
$\mathrm{v} = 1 0 - (2) (2) = 6 \mathrm{m} / \mathrm{s}$

Retardation remains same in both situations $\frac{0 - (200)^2}{2\times 4} = \frac{0 - V^2}{2\times 9}$ $V = 300~\mathrm{cm / s}$

A

Time of free fall does not depend on mass.

$\mathrm{v}^2 = 2\mathrm{as}$ $\mathrm{v} = \sqrt{2\mathrm{gh}}$ [does not depend upon mass]

$\mathrm{h} = \frac{1}{2}\mathrm{gt}^2 = \frac{1}{2}\times 10\times 5^2 = 125\mathrm{m}$

Let total time of fall = n, u = 0 $S_{last} = S_{3}$ $\frac{g}{2}(2n-1)=\frac{1}{2}g(3)^{2}$ 2n - 1 = 9 $\Rightarrow$ t = 5 sec

$\mathrm{x} = \frac{1}{2}\mathrm{g}(1)^2 = \frac{1}{2}\mathrm{g}$ $5\mathrm{x} = \frac{\mathrm{g}}{2}(2\mathrm{n} - 1)$ $5\times \frac{\mathrm{g}}{2} = \frac{\mathrm{g}}{2}(2\mathrm{n} - 1)$ n = 3 sec.

$\mathbf {v} ^ {2} = \mathbf {u} ^ {2} + 2 \mathrm{as}$
$\mathbf {v} ^ {2} = 2 \mathbf {g} (2 \mathbf {d})$
$\mathrm{v} = 2 \sqrt {\mathrm{gd}}$

use $v^{2}=u^{2}-2as$ $\Rightarrow0=u^{2}-2as$ $\Rightarrow a=\frac{u^{2}}{2s}$

Distance covered in 1st sec is $x = \frac{1}{2} g(1)^2 = \frac{g}{2}$ Let it falls for $n$ sec then last sec is $n^{\text{th}}$ sec. $\therefore 9x = 0 + \frac{g}{2} (2n - 1)$ $\Rightarrow 9 \times \frac{g}{2} = \frac{g}{2} (2n - 1)$ $\Rightarrow 2n = 10$ $\Rightarrow n = 5 \sec$

Time taken by stone to go down $t_{1}=\sqrt{\frac{2h}{g}}$ Time taken by sound to come up $t_{2}=\frac{h}{v}$ $\therefore T=t_{1}+t_{2}=\sqrt{\frac{2h}{g}}+\frac{h}{v}$

$H=\frac{1}{2}gt^{2}$ $h=\frac{1}{2}g\left(\frac{t}{3}\right)^{2}=\frac{H}{9}$ from releasing point so distance from the ground $=H-H/9=8H/9$

$h=0+\frac{1}{2}gt_{1}^{2}$ $\Rightarrow t_{1}=\sqrt{\frac{2h}{g}}$
$\frac {\mathrm{h}}{4} = 0 + \frac {1}{2} \mathrm{gt} _ {2} ^ {2} \Rightarrow \mathrm{t} _ {2} ^ {2} = \frac {2}{\mathrm{g}} \times \frac {\mathrm{h}}{4}$
$\Rightarrow \quad t _ {2} = \sqrt {\frac {2}{g} \times \frac {h}{4}} = \frac {1}{2} \sqrt {\frac {2 h}{g}} = \frac {t _ {1}}{2}$

Total height to fall
$\mathrm{h} = \frac {1}{2} \mathrm{gt} ^ {2} = \frac {1}{2} \times 1 0 \times (5) ^ {2} = 1 2 5 \mathrm{m}$
In first 3 sec.
$s _ {1} = \frac {1}{2} g t _ {1} ^ {2} = \frac {1}{2} \times 1 0 \times (3) ^ {2} = 4 5 m$
Now remaining height to fall
$\therefore \quad s _ {2} = h - s _ {1} = 1 2 5 - 4 5 = 8 0 m$
$t _ {2} = \sqrt {\frac {2 s _ {2}}{g}} = \sqrt {\frac {2 \times 8 0}{1 0}} = 4 \sec$
$\mathrm{t} = \mathrm{t} _ {1} + \mathrm{t} _ {2} = 3 + 4 = 7 \sec$

Separation $= S_{1} - S_{2}$ $= \frac{1}{2}\mathrm{g}(4)^{2} - \frac{1}{2}\mathrm{g}(2)^{2}$ $= \frac{12}{2}\mathrm{g} = 58.8\mathrm{m}$

$\mathrm{h}_1 - \mathrm{h}_2 = \frac{1}{2}\mathrm{g}(\mathrm{t}_1^2 -\mathrm{t}_2^2)$ $= \frac{1}{2}\times 9.8[(5)^2 -(3)^2 ]$ $= 8\times 9.8 = 78.4\mathrm{m}$

In first 2 sec the distance travelled is $h_{1}=0+\frac{1}{2}(10)(2)^{2}=20\ m$ After stopped & released the distance travelled
in next 2 sec $h_{2}=0+\frac{1}{2}10(2)^{2}=20\ m$ So height from ground = (H - 40)m

$S_{\mathrm{n^{th}}} = u + \frac{a}{2} (2n - 1)$
$\mathrm{u} = 0 \text { and } \mathrm{a} = \mathrm{g(const.)}$
$S _ {n ^ {\text {th}}} \propto (2 n - 1)$
$(\text {here n} = 1, 2, 3.....)$
$D_{1}:D_{2}:D_{3}:\ldots\ldots=1:3:5:\ldots\ldots$

$\because \mathrm{h} = \frac{1}{2}\mathrm{gt}^2$
∴ $t = \sqrt{\frac{2h}{g}}$
$\begin{array}{l} \left[ \begin{array}{c} t = 0 \\ t _ {1} = \sqrt {\frac {2}{g}} \sqrt {1} \end{array} \right] - t _ {1 k m} = \sqrt {\frac {2}{g}} \sqrt {1} \\ \left[ \begin{array}{c} t _ {2} = \sqrt {\frac {2}{g}} \sqrt {2} \end{array} \right] - t _ {I I k m} = \sqrt {\frac {2}{g}} (\sqrt {2} - \sqrt {1}) \\ \left[ \begin{array}{c} t _ {3} = \sqrt {\frac {2}{g}} \sqrt {3} \end{array} \right] - t _ {I I I k m} = \sqrt {\frac {2}{g}} (\sqrt {3} - \sqrt {2}) \\ \left[ \begin{array}{c} t _ {4} = \sqrt {\frac {2}{g}} \sqrt {4} \end{array} \right] - t _ {I V k m} = \sqrt {\frac {2}{g}} (\sqrt {4} - \sqrt {3}) \end{array}$
$\begin{array}{r l} & \therefore t _ {I k m}; t _ {I I k m}; t _ {I I I k m}; \dots \\ & = \sqrt {1}: (\sqrt {2} - \sqrt {1}): (\sqrt {3} - \sqrt {2}): \dots \end{array}$

By galileo's law of odd number
$x + 3x + 5x = 27m$ $x = 3 m$ Position of $2^{nd}$ drop from top $x + 3x = 4x = 12m$ Position of $3^{rd}$ drop from top
x = 3 m

By galileo's law of odd number
$\mathrm{x} + 3 \mathrm{x} = 6 \mathrm{m}$
4 x = 6
$\mathrm{x} = \frac {6}{4} = 1. 5 \mathrm{m}$
Height of $2^{\mathrm{nd}}$ drop from ground
$= 3 \mathrm{x} = 3 \times \frac {6}{4} = 4. 5 \mathrm{m}$

$S_{1} = \frac{1}{2} g(4)^{2}$
$\mathrm{S} _ {1} = 8 0 \mathrm{m}$
$\mathrm{S} _ {2} = \frac {1}{2} \mathrm{g} (3) ^ {2} = 4 5 \mathrm{m}$
$\mathrm{S} _ {3} = \frac {1}{2} \mathrm{g} (2) ^ {2} = 2 0 \mathrm{m}$
$\mathrm{S} _ {4} = \frac {1}{2} \mathrm{g} (1) ^ {2} = 5 \mathrm{m}$
$\mathrm{S} _ {1} - \mathrm{S} _ {2} = 8 0 - 4 5 = 3 5 \mathrm{m}$
$\mathrm{S} _ {2} - \mathrm{S} _ {3} = 4 5 - 2 0 = 2 5 \mathrm{m}$
$\mathrm{S} _ {3} - \mathrm{S} _ {4} = 2 0 - 5 = 1 5 \mathrm{m}$

$\mathrm{v} ^ {2} = \mathrm{u} ^ {2} - 2 \mathrm{g} \frac {\mathrm{h}}{2}$
$0 = (1 0) ^ {2} - 1 0 \mathrm{h}$
$\mathrm{h} = 1 0 \mathrm{m}$