Kinematics-1D

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
$= 3 \mathrm{x} = 3 \times \frac {6}{4} = 4. 5 \mathrm{m}$

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

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

$\mathrm{T} = \frac{2\mathrm{u}}{\mathrm{g}} = \frac{2\times 50}{10} = 10\mathrm{sec}$

$\because$ Height $H = \frac{u^2}{2g}$
$\mathrm{H} \propto \mathrm{u} ^ {2}$
$\frac {\mathrm{H} _ {1}}{\mathrm{H} _ {2}} = \frac {(2 \mathrm{u}) ^ {2}}{(3 \mathrm{u}) ^ {2}} = \frac {4}{9}$
$\because \text { Time } T = \frac {2 u}{g}$
$\mathrm{T} \propto \mathrm{u}$
$\frac {\mathrm{T} _ {1}}{\mathrm{T} _ {2}} = \left(\frac {2 \mathrm{u}}{3 \mathrm{u}}\right) = \frac {2}{3}$

$\mathrm{H}_{\mathrm{max.}} = \frac{\mathrm{u}^2}{2\mathrm{g}} = \frac{40\times 40}{2\times 10}$
$\mathrm{H} _ {\max.} = 8 0 \mathrm{m}$

v = u + at,
$\mathrm{v} = 4 0 - 1 0 (3) = 1 0 \mathrm{m/s}.$

$\frac{\mathrm{h}_2}{\mathrm{h}_1} = \frac{\frac{\mathrm{u}_2}{2\mathrm{g}}}{\frac{\mathrm{u}_1^2}{2\mathrm{g}}} \Rightarrow \frac{2\mathrm{h}_1}{\mathrm{h}_1} = \frac{\mathrm{u}_2^2}{\mathrm{v}_0^2}$
$\mathrm{u} _ {2} ^ {2} = 2 \mathrm{v} _ {0} ^ {2} \Rightarrow \mathrm{u} _ {2} = \sqrt {2} \mathrm{v} _ {0}$

It is only possible when $7^{\text{th}}$ sec is the last
second for its ascending motion and $8^{\text{th}}$ sec is first second of its descending motion, which is clear from figure.

∴ 0 = u - gt
so u = gt = 9.8 × 7 = 68.6 m/sec

Distance covered in last t second of ascent
= distance covered in first t second of descent $= 0 + \frac{1}{2}gt^{2} = \frac{1}{2}gt^{2}$

Distance travelled in $1^{st}$ second during upward motion = distance travelled in last second during downward motion.
$S _ {1 ^ {\mathrm{st}}} = \frac {\mathrm{g}}{2} (2 \times 6 - 1) = \frac {1 1 \mathrm{g}}{2}$
$S _ {7 ^ {\mathrm{th}}} = \frac {1}{2} \mathbf {g} (1) ^ {2} = \frac {\mathbf {g}}{2}$
$\frac {S _ {1 ^ {\mathrm{st}}}}{S _ {7 ^ {\mathrm{th}}}} = \frac {1 1}{1}$

$S=\frac{1}{2}gt^{2}$
$\mathrm{S} = \frac {1}{2} \mathrm{g} (2) ^ {2} = 2 0 \mathrm{m}$

$S_{1} = S_{2}$
$9 8 (t) - \frac {1}{2} (9. 8) t ^ {2} = 9 8 (t - 4) - \frac {1}{2} (9. 8) (t - 4) ^ {2}$
$1 0 t - \frac {t ^ {2}}{2} = 1 0 (t - 4) - \frac {1}{2} (t - 4) ^ {2}$
$\Rightarrow 4 t - 4 8 = 0$
$\Rightarrow \quad t = \frac {4 8}{4} = 1 2 s$

For distance
'S' time should be $\Delta t/2$
$S = \frac {1}{2} \mathrm{g} \left(\frac {\Delta t}{2}\right) ^ {2}$
For total height
$(h + S) = \frac {u ^ {2}}{2 g}$
$\mathrm{u} = \sqrt {2 \mathrm{g} (\mathrm{h} + S)}$
$= \sqrt {2 g \left(h + \frac {1}{2} g \left(\frac {\Delta t}{2}\right) ^ {2}\right)}$
$= \sqrt {2 \mathrm{gh} + \left(\frac {\mathrm{g} \Delta t}{2}\right) ^ {2}}$

Let retardation due to
then for upward journey
net retardation = (g + a)
$t _ {\mathrm{asc}} = \sqrt {\frac {2 h}{g + a}}$
and for downward journey
net acceleration = (g - a)
$\therefore \quad t _ {\mathrm{desc}} = \sqrt {\frac {2 h}{g - a}}$
\therefore \quad t _ {\text {asc}} < t _ {\text {desc}}

There is loss of kinetic energy during the motion so it is clear that $v_{1} > v_{2}$

$\frac{t_{ascent}}{t_{descent}}=\sqrt{\frac{g-a}{g+a}}=\sqrt{\frac{10-2}{10+2}}=\sqrt{\frac{2}{3}}$

$\because \mathrm{v}^2 = \mathrm{u}^2 + 2\mathrm{g(H)}$
$\mathrm{v} ^ {2} = (2 0) ^ {2} + 2 (9. 8) (2 0 0)$
$\mathbf {v} = \sqrt {4 3 2 0}$
$\boxed {\mathrm{v} \simeq 6 5 \mathrm{m/s}}$