Kinematics-2D

$v = \sqrt{2gR\sin\alpha}$
$N - mg\sin \alpha = \frac{mv^2}{R} = 2mg\sin \alpha$
$\Rightarrow N = 3mg\sin \alpha$


$\therefore A = \frac {\left(\frac {m v ^ {2}}{R}\right)}{N} = \frac {2 m g \sin \alpha}{3 m g \sin \alpha} = \frac {2}{3}.$

$N = \frac{mv^2}{r}$
Curve is parabola
$Y = k x ^ {2}.$

(I) $v_{rel} = \sqrt{v^{2} + v^{2}} = \sqrt{2}v = \sqrt{2}\omega R = \sqrt{2}$

$\begin{array}{r l} & {\vec {v} _ {A} = v \cos 4 5 ^ {\circ} (\hat {i}) + (v \sin 4 5 ^ {\circ} - g t) \hat {j}} \\ & {\vec {v} _ {B} = v \cos 4 5 ^ {\circ} (- \hat {i}) + [ v \sin 4 5 ^ {\circ} - g (t - 0. 1) ] \hat {j}} \\ & {\vec {v} _ {A} - \vec {v} _ {B} = 2 v \cos 4 5 ^ {\circ} \hat {i} - g (0. 1) \hat {j}} \\ & {= \frac {2 v}{\sqrt {2}} \hat {i} - \hat {j} = \sqrt {2} \frac {5 \pi}{\sqrt {2}} \hat {i} - \hat {j} = 5 \pi \hat {i} - \hat {j}} \\ & {| \vec {v} _ {B} - \vec {v} _ {A} | = \sqrt {2 5 \pi^ {2} + 1}.} \end{array}$
$v _ {A} = \frac {x _ {0}}{t _ {0}} \cos \frac {t}{t _ {0}} = 1 \cos \frac {\pi}{3} = \frac {1}{2}$
$v _ {B} = \frac {x _ {0}}{t _ {0}} \cos \left(\frac {t}{t _ {0}} + \frac {\pi}{2}\right) = 1 \cos \left(\frac {\pi}{3} + \frac {\pi}{2}\right)$
$= - \sin \frac {\pi}{3} = \frac {- \sqrt {3}}{2}$
$v _ {r e l} = \frac {\sqrt {3} + 1}{2}$
(IV) $v_{rel} = \sqrt{3^2 + 1^2} = \sqrt{10}$ .

$y = \alpha x - \beta x^2$
Comparing with trajectory equation
$y = x \tan \theta - \frac {1}{2} \frac {g x ^ {2}}{u ^ {2} \cos^ {2} \theta}$
$\tan \theta = \alpha \Rightarrow \theta = \tan^ {- 1} \alpha$
$\beta = \frac {1}{2} \frac {g}{u ^ {2} \cos^ {2} \theta}$
Maximum height : H
$H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \frac {g}{2 \beta \cos^ {2} \theta} \frac {\sin^ {2} \theta}{2 g}$
$H = \frac {\tan^ {2} \theta}{4 \beta} = \frac {\alpha^ {2}}{4 \beta}.$

Statement-I :
$\begin{array}{l} v _ {\max} = \sqrt {\mu R g} = \sqrt {(0 . 2) \times 2 \times 9 . 8} = 1. 9 7 m / s \\ 7 k m / h = 1. 9 4 4 m / s \end{array}$
Speed is lower than $v_{max}$ , hence it can take safe turn.
Statement-II :
$v _ {\max} = \sqrt {R g \left[ \frac {\tan \theta + \mu}{1 - \mu \tan \theta} \right]}$
$= \sqrt {2 \times 9 . 8 \left[ \frac {1 + 0 . 2}{1 - 0 . 2} \right]} = 5. 4 2 m / s$

$\mu_sN=\frac{mv^2}{R}$
$N=\frac{mv^2}{\mu_sR}=mg+F_L$
$F_L=\frac{mv^2}{\mu_sR}-mg=m\left(\frac{v^2}{\mu_sR}-g\right)$

$V_{BW}=$ velocity of butterfly in frame of wind
$\vec{V}_{BW}=4\sqrt{2}\cos45^\circ\hat{i}+4\sqrt{2}\sin45^\circ\hat{j}$
$\vec{V}_{BW}=4\hat{i}+4\hat{j}$
$\vec{V}_{W}=-\hat{j}$
$\vec{V}*{B}=\vec{V}*{BW}+\vec{V}_{W}=4\hat{i}+3\hat{j}$
$\vec{V}*{B/A}=\vec{V}*{B}-\vec{V}_{A}=(4\hat{i}+3\hat{j})-(3\hat{i}-6\hat{j})$
$=1\hat{i}+9\hat{j}$
$\therefore |S_B|=\sqrt{(12)^2+(9)^2}=15,m$

$42^\circ+48^\circ=90^\circ$
Then range will be same $R_1=R_2$
$\theta_1<\theta_2$
So, $H_1 i.e. for maximum $\theta$, height will be maximum.

Horizontal range of food packet $=v\times\sqrt{\frac{2h}{g}}$
Also, horizontal distance travelled by aeroplane
$=v\times\sqrt{\frac{2h}{g}}$
Distance between aeroplane and person $=h$
$\therefore$ distance $=\sqrt{\frac{2h}{g}v^2+h^2}$

$\theta=45^\circ$
$H=\frac{u^2\sin^2\theta}{2g}
$=\frac{(25)^2\times\left(\frac{1}{\sqrt2}\right)^2}{2\times10}$
=\frac{125}{8},m=15.625,m$
and time $t=\frac{T}{2}$
$=\frac{u\sin\theta}{g}
$=\frac{25\times\frac{1}{\sqrt2}}{10}$
$=\frac{25}{10\sqrt2}$
=\frac{5}{2\sqrt2},s=1.77,s$

$\sin\theta=\frac{1}{\sqrt2}$
$\theta=45^\circ$
$\tan\theta=\frac{\frac{mv^2}{r}}{mg}=\frac{v^2}{rg}=1$
$v^2=rg\Rightarrow v=\sqrt{rg}$

$U = -\frac{C}{r}$
$F = - \frac {d U}{d r} = - \frac {C}{r ^ {2}}$
$| F | = \frac {m v ^ {2}}{r} \Rightarrow \frac {C}{r ^ {2}} = \frac {m v ^ {2}}{r} \Rightarrow v ^ {2} \propto \frac {1}{r}.$

$x = u_{x}t + \frac{1}{2} a_{x}t^{2}, y = u_{y}t + \frac{1}{2} a_{y}t^{2}$
$32 = 0 \times t + \frac {1}{2} (4) (t) ^ {2} \Rightarrow t ^ {2} = 1 6 \Rightarrow t = 4 \sec$ $x = 3 \times 4 + \frac {1}{2} \times 6 \times 4 ^ {2} = 1 2 + 4 8 = 6 0 m.$

$F_{net} = ma$
$- m g - m k v ^ {2} = m v \frac {d v}{d s}$
$v \frac {d v}{d s} = - g - k v ^ {2}$
$- \int_ {u} ^ {0} \frac {v d v}{g + k v ^ {2}} = \int_ {0} ^ {h _ {\max}} d s = h _ {\max}$
$h _ {\max} = \frac {1}{2 k} \ln \left(\frac {g + k u ^ {2}}{g}\right) = \frac {1}{2 k} \ln \left(1 + \frac {k u ^ {2}}{g}\right).$

$R = 0.1m$
$\omega = \frac {2 \pi}{T} = \frac {2 \pi}{6 0} = 0. 1 0 5 r a d / \sec$
$a = \omega^ {2} R = (0. 1 0 5) ^ {2} (0. 1) = 0. 0 0 1 1 = 1. 1 \times 1 0 ^ {- 3}.$
Average acceleration is of the order of $10^{-3}$ .

$\frac{dv_x}{dt} = \frac{k}{m} v_y, \frac{dv_y}{dt} = \frac{k}{m} v_x$
$\frac {d v _ {y}}{d v _ {x}} = \frac {v _ {x}}{v _ {y}} \Rightarrow \int v _ {y} d v _ {y} = \int v _ {x} d v _ {x}$
$v _ {y} ^ {2} = v _ {x} ^ {2} + C \Rightarrow v _ {y} ^ {2} - v _ {x} ^ {2} = \text { constant }$
Option (c)
$\vec {v} \times \vec {a} = (v _ {x} \hat {i} + v _ {y} \hat {j}) \times \frac {k}{m} (v _ {y} \hat {i} + v _ {x} \hat {j})$
$= (v _ {x} ^ {2} \hat {k} - v _ {y} ^ {2} \hat {k}) \frac {k}{m} = (v _ {x} ^ {2} - v _ {y} ^ {2}) \frac {k}{m} \hat {k} = \text { constant }.$

Velocity at ground (means zero height) is non-zero therefore one is incorrect and velocity versus height is non-linear therefore two is also incorrect
$v ^ {2} = 2 g h, v \frac {d v}{d h} = 2 g = \text { constant }$
$\frac {d v}{d h} = \frac {\text { constant }}{v}$
Here we can see slope is very high when velocity is low therefore at Maximum height the slope should be very large which is in option c) and as velocity increases slope must decrease there for option c) is correct.

$1 5 = 1 0 \times \frac {1}{2} \sqrt {3} - \frac {1}{2} a (\sqrt {3}) ^ {2}$
$\frac {3}{2} a = 5 \sqrt {3} - 1. 1 5$
$\frac {3 a}{2} = 8. 6 5 - 1. 1 5 = 7. 5$
$a = 7. 5 \times \frac {2}{3} \approx 5 \mathrm{m} / \sec^ {2} \Rightarrow a = 5 \mathrm{m} / \sec^ {2}.$

$\vec{r} = A\cos \omega t\hat{i} +A\sin \omega t\hat{j} +A\omega t\hat{k}$ $\vec{v} = A\omega \sin \omega t\hat{i} +A\omega \cos \omega t\hat{j} +A\omega \hat{k}$ $|\vec{v}| = \omega A\sqrt{(-\sin\omega t)^2 + (\cos\omega t)^2 + (1)^2} = \sqrt{2}\omega A.$

$R = \frac{u^2\sin 2\theta}{g}$
∵ Max. Range= $\frac{u^{2}}{g}$
$A = \pi R ^ {2}$
$A \propto R ^ {2} \Rightarrow A \propto u ^ {4}$
${\frac {A _ {1}}{A _ {2}}} = {\frac {u _ {1} ^ {4}}{u _ {2} ^ {4}}} = \left[ {\frac {1}{2}} \right] ^ {4} = {\frac {1}{1 6}}.$

$u _ {x} = 1 0 \cos 6 0 ^ {\circ} = 5 m / s$
$u _ {y} = 1 0 \cos 3 0 ^ {\circ} = 5 \sqrt {3} m / s$
$\text { velocity after } t = 1 \text { sec. }$
$v _ {x} = 5 m / s$
$v _ {y} = \left| 5 \sqrt {3} - 1 0 \right| m / s = 1 0 - 5 \sqrt {3} m / s$
$= \sqrt {2 5 + 7 5 + 1 0 0 - 1 0 0 \sqrt {3}} = \sqrt {2 0 0 - 1 0 0 \sqrt {3}} = 5. 1 1 m / s$
$r _ {c} = \frac {v ^ {2}}{g \cos \theta} = \frac {v ^ {2}}{g \frac {u _ {x}}{v}} = \frac {v ^ {3}}{g u _ {x}} = \frac {(5 . 1 1) ^ {3}}{1 0 \times 5} = 2. 8 m.$

$\mid \Delta \overline {{v}} \mid = \sqrt {v _ {1} ^ {2} + v _ {2} ^ {2} + 2 v _ {1} v _ {2} \cos (\pi - \theta)}$
$= 2 v \sin {\frac {\theta}{2}}$
since $\left[\mid \overline{v}_1\mid = \mid \overline{v}_2\mid \right]$
$= (2 \times 1 0) \times \sin (3 0 ^ {\circ}) = 1 0 m / s.$