Kinematics-2D

$\begin{array}{l} H = \frac {u ^ {2} \cos^ {2} \beta}{2 g} \\ \Rightarrow u \cos \beta = \sqrt {2 g H} \end{array}$
$\mathrm{So}, t = \frac {T}{2} = \frac {u \cos \beta}{g} = \sqrt {\frac {2 H}{g}}$
Hence, the correct answer is (B).

Since, time of flight does not depend on horizontal acceleration, so
$\begin{array}{r l} & T = \frac {2 u \sin \theta}{g} = \frac {2 (2 0) \left(\frac {1}{\sqrt {2}}\right)}{1 0} = 2 \sqrt {2} \mathrm{s} \\ \text {Since} & x = u _ {x} t + \frac {1}{2} a _ {x} t ^ {2} \\ & \Rightarrow R = u _ {x} T + \frac {1}{2} a _ {x} T ^ {2} \\ & \Rightarrow R = (2 0 \cos 4 5 ^ {\circ}) (2 \sqrt {2}) + \frac {1}{2} (2) (2 \sqrt {2}) ^ {2} \\ & \Rightarrow R = 4 0 + 8 = 4 8 \mathrm{m} \end{array}$
Hence, the correct answer is (B).

Since, particle hits the inclined plane at an angle of $30^{\circ}$ , which means that the particle hits the plane horizontally. So, this is the highest point of trajectory.
$\Rightarrow t = \frac {T}{2} = \frac {u \sin \theta}{g} = \frac {2 0 \sin (6 0 ^ {\circ})}{g}$
Hence, the correct answer is (A).

$\begin{array}{l} \text {Since, v_{x} = \frac {dx}{dt} = 2 ms^{-1}} \\ \Rightarrow x = 2 t \\ \text {Also, v_{y} = \frac {dy}{dt} = 5x} \\ \Rightarrow d y = x d t = 5 (2 t) d t \\ \Rightarrow d y = 1 0 t d t \\ \Rightarrow y = \frac {1 0 t ^ {2}}{2} = 5 t ^ {2} \end{array}\tag{... (1}$
{from (1)}
$\ldots(2)$
From (1) and (2), we get
$\begin{array}{r l} & y = 5 \left(\frac {x}{2}\right) ^ {2} \\ \Rightarrow & y = \frac {5}{4} x ^ {2} \end{array}$
Hence, the correct answer is (A).

Since, $\frac{u^2\sin^2\theta}{2g} = 5\mathrm{m}$ $\Rightarrow \frac{(20)^2\sin^2\theta}{20} = 5$ $\Rightarrow \sin \theta = \frac{1}{2}$ $\Rightarrow \theta = 30^{\circ}$ So, $R = \frac{u^2\sin(2\theta)}{g} = \frac{(20)^2\sin(60^{\circ})}{10} = 20\sqrt{3}$
Hence, the correct answer is (D).

C

Since, $y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$ $\Rightarrow h = d \tan \theta - \frac{gd^2}{2u^2 \cos^2 \theta}$ $\Rightarrow \frac{gd^2}{2u^2 \cos^2 \theta} = (d \tan \theta - h)$ $\Rightarrow u^2 = \frac{gd^2}{2 \cos^2 \theta (d \tan \theta - h)}$ $\Rightarrow u = \frac{d}{\cos \theta} \sqrt{\frac{g}{2(d \tan \theta - h)}}$ Hence, the correct answer is (
Hence, the correct answer is (B).

Apply formula of time of flight on inclined plane. Hence, the correct answer is (B).

$T = \frac{2u \sin \theta}{g}$ $\Rightarrow R = \frac{u^{2} \sin(2\theta)}{g} - \frac{2u^{2} \sin^{2} \theta}{g}$ R will be maximum when, $\frac{dR}{d\theta} = 0$ $\Rightarrow 2 \cos(2\theta) - 2 \sin(2\theta) = 0$ $\Rightarrow \tan(2\theta) = 1$ $\Rightarrow 2\theta = \frac{\pi}{4}$ $\Rightarrow \theta = \frac{\pi}{8}$ radian
Hence, the correct answer is (B).

$u = 20\mathrm{ms}^{-1},\theta = 45^{\circ}$ Given $h = Ax - Bx^{2}$ Comparing with, $y = x\tan \theta -\frac{gx^2}{2u^2\cos^2\theta}$
We get $A = \tan\theta = \tan45^{\circ} = 1$ and $B = \frac{b}{2u^{2}\cos^{2}\theta} = \frac{10}{2(20)^{2}\left(\frac{1}{2}\right)} = \frac{1}{40}$ $\Rightarrow \frac{A}{B}=\frac{40}{1}$
Hence, the correct answer is (D).

$\begin{array}{l} v _ {x} = \frac {d x}{d t} = 2 + 8 t \\ \Rightarrow a _ {x} = \frac {d v _ {x}}{d t} = 8 \\ v _ {y} = \frac {d y}{d t} = 5 \\ \Rightarrow a _ {y} = \frac {d v _ {y}}{d t} = 0 \end{array}$
Hence, the correct answer is (C).

$v_{x}=\frac{dx}{dt}=v_{0}$ $\Rightarrow x=v_{0}t$ Since, $v_{y}=\frac{dy}{dt}=a\omega\cos(\omega t)$ $\Rightarrow y=\int_{0}^{t}a\omega\cos(\omega t)dt=\frac{a\omega\sin(\omega t)}{\omega}$ $\Rightarrow y=a\sin(\omega t)=a\sin\left(\frac{\omega x}{v_{0}}\right)$ Hence, the correct answer is (C).

$\begin{array}{l} \frac {\frac {1}{2} m u _ {1} ^ {2} \cos^ {2} \theta_ {1}}{\frac {1}{2} m u _ {2} ^ {2} \cos^ {2} \theta_ {2}} = \frac {4}{1} \\ \Rightarrow \quad \frac {u _ {1} \cos \theta_ {1}}{u _ {2} \cos \theta_ {2}} = 2 \\ \text { and } \quad \frac {u _ {1} ^ {2} \sin^ {2} \theta_ {1}}{u _ {2} ^ {2} \sin^ {2} \theta_ {2}} = \frac {4}{1} \\ \Rightarrow \quad \frac {u _ {1} \sin \theta_ {1}}{u _ {2} \sin \theta_ {2}} = \frac {2}{1} \end{array}$
From equations (1) and (2), we get
$\frac {(u _ {1} \sin \theta_ {1}) (u _ {1} \cos \theta_ {1})}{(u _ {2} \sin \theta_ {2}) (u _ {2} \cos \theta_ {2})} = \frac {4}{1}$
$\Rightarrow \quad \frac {\frac {g R _ {1}}{2}}{\frac {g R _ {2}}{2}} = \frac {4}{1}$
$\Rightarrow \frac {R _ {1}}{R _ {2}} = \frac {4}{1}$
Hence, the correct answer is (B).

$\begin{array}{l} {2 = \sqrt {u ^ {2} \sin^ {2} \theta - 2 h g}} \\ {\Rightarrow \quad 4 = u ^ {2} \sin^ {2} \theta - 2 (0. 4) (1 0)} \end{array}$

$\begin{array}{l l} \Rightarrow & u ^ {2} \sin^ {2} \theta = 1 2 \\ \Rightarrow & u \sin \theta = 2 \sqrt {3} \text {and} u \cos \theta = 6 \\ \Rightarrow & \tan \theta = \frac {1}{\sqrt {3}} \\ \Rightarrow & \theta = 3 0 ^ {\circ} \end{array}$
Hence, the correct answer is (C).

$\begin{array}{l} u _ {x} = 1 4. 7 \mathrm{ms} ^ {- 1}, \\ u _ {y} = 9. 8 \times 2 = 1 9. 6 \mathrm{ms} ^ {- 1} \\ \Rightarrow u = \sqrt {(1 4 . 7) ^ {2} + (1 9 . 6) ^ {2}} = 2 4. 5 \mathrm{ms} ^ {- 1} \\ \Rightarrow \tan \theta = \frac {1 9 . 6}{1 4 . 7} = \frac {4}{3} \\ \Rightarrow \theta = 5 3 ^ {\circ} \\ \text { Hence,the correct answer is(D). } \end{array}$

Since, both projectiles have equal height, so
$H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} = \frac {u _ {y} ^ {2}}{2 g}$
So, time of flight of both will be same.
$\Rightarrow T _ {1} = T _ {2} = \frac {2 u \sin \theta}{g} = \frac {2 u _ {y}}{g}$
Since Range is greater for second, i.e.
$R _ {2} > R _ {1}$
Thus $\left(u_x\right)_2 > \left(u_x\right)_1$
$\Rightarrow u _ {2} > u _ {1}$
Hence, the correct answer is (D).

Acceleration of the bead down the wire is $g \cos \theta$ .
Hence $v^{2}=0^{2}+2(g\cos\theta)(2R\cos\theta)$ Where $P_{1}P_{2}=2R\cos\theta$ $\Rightarrow v=2\left(\sqrt{gR}\right)\cos\theta$ Since, $t=\frac{v}{a}=\frac{2\sqrt{gR}\cos\theta}{g\cos\theta}=2\sqrt{\frac{R}{g}}$
Hence, (B), (C) and (D) are correct.

$y = x \tan \theta \left(1 - \frac {x}{R}\right)$
where $x = (u\cos \theta)t$ and $R = (u\cos \theta)T$
$\ldots(2)$
Also, we observe that
$\begin{array}{l l} & R = \frac {2 u ^ {2} \sin \theta \cos \theta}{g} \text {and} H = \frac {u ^ {2} \sin^ {2} \theta}{2 g} \\ \Rightarrow & \frac {H}{R} = \frac {\sin^ {2} \theta}{4 \sin \theta \cos \theta} = \frac {1}{4} \tan \theta \\ \Rightarrow & \tan \theta = \frac {4 H}{R} \end{array}\tag{... (3}$
So, we get
$y = \frac {4 H x}{R} \left(1 - \frac {x}{R}\right)$
Substitute x and R from (2), we get
$y = 4 H \left(\frac {t}{T}\right) \left(1 - \frac {t}{T}\right)$
Hence, (A) and (C) are correct.

$\vec {u} = (u \cos \theta) \hat {i} + (u \sin \theta) \hat {j}\tag{... (1}$
Let the velocity at point Q, at time t be $\vec{v}$ , then
$\vec {v} = (u \cos \theta) \hat {i} + (u \sin \theta - g t) \hat {j}$
Since $\vec{v} \perp \vec{u}$ , so
$\begin{array}{r l} & {\vec {v} \cdot \vec {u} = 0} \\ {\Rightarrow} & {u ^ {2} - u g t \sin \theta = 0} \end{array}$
$\Rightarrow t = \frac {u}{g \sin \theta} = \left(\frac {u}{g}\right) \mathrm{cosec} \theta$
Speed of the particle at Q is $|\vec{v}|$ , where
$| \vec {v} | = \sqrt {u ^ {2} + g ^ {2} t ^ {2} - 2 u g t \sin \theta}$
Substituting the value of t, we get
$| \vec {v} | = u \cot \theta$
Hence, (C) and (D) are correct.

$\begin{array}{l l} & h = 2 \sqrt {h g} (\sin \theta) t - \frac {1}{2} g t ^ {2} \\ \Rightarrow & t ^ {2} - 4 \sqrt {\frac {h}{g}} (\sin \theta) t + \frac {2 h}{g} = 0 \\ \Rightarrow & t = \frac {4 \sqrt {\frac {h}{g}} \pm \sqrt {1 6 \left(\frac {h}{g}\right) \sin^ {2} \theta - \frac {8 h}{g}}}{2} \\ \Rightarrow & t = \frac {4 \sqrt {\frac {h}{g}} \pm \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & t _ {1} = \frac {4 \sqrt {\frac {h}{g}} - \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & t _ {2} = \frac {4 \sqrt {\frac {h}{g}} + \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1}}{2} \\ \Rightarrow & (t _ {2} - t _ {1}) = \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1} \end{array}$
Further horizontal distance travelled during this time is 2h
$\begin{array}{l l} \Rightarrow & 2 h = 2 \sqrt {h g} \cos \theta (t _ {2} - t _ {1}) \\ \Rightarrow & 2 h = 2 \sqrt {h g} \cos \theta \sqrt {\frac {8 h}{g}} \sqrt {2 \sin^ {2} \theta - 1} \\ & \text {Solving we get} \cos \theta = \frac {1}{2} \end{array}$
$\theta = 60^{\circ}$ with horizontal or $30^{\circ}$ with vertical
$\Rightarrow \quad (t _ {2} - t _ {1}) = 2 \sqrt {\frac {h}{g}}$
Hence, (A) and (D) are correct.

$\begin{aligned} &\vec{u} = 3\hat{i}\text{ m/s} \quad \text{and} \quad \vec{a} = -\hat{i} - 0.5\hat{j}\text{ m/s}^2 \\ &\Rightarrow u_x = 3\text{ ms}^{-1}, \ a_x = -1\text{ ms}^{-2} \quad \text{and} \quad a_y = -0.5\text{ ms}^{-2} \end{aligned}$
Since $u_x$ and $a_x$ have opposite signs, the particle must first reverse its direction of motion along the x-axis. Let this happen when the x-coordinate of the particle is $x$. Then:
$\begin{aligned} & 0^2 - u_x^2 = 2 a_x x \\ \Rightarrow & -(3)^2 = 2 (-1) x \\ \Rightarrow & x = 4.5\text{ m} \end{aligned}$
Also, if this happens at time $t$, then:
$\begin{aligned} & v_x = u_x + a_x t \quad \text{gives} \quad 0 = 3 + (-1) t \\ \Rightarrow & t = 3\text{ s} \end{aligned}$
At this time $t$, the y-coordinate of the particle is given by:
$\begin{aligned} & y = u_y t + \frac{1}{2} a_y t^2 \\ \Rightarrow & y = 0 + \frac{1}{2} (-0.5) (3)^2 \\ \Rightarrow & y = -\frac{9}{4}\text{ m} = -2.25\text{ m} \\ \text{Also}, \quad & v_y = 0 + a_y t = -(0.5) (3) = -1.5\text{ ms}^{-1} \\ \Rightarrow & \vec{v} = v_x \hat{i} + v_y \hat{j} = 0 \hat{i} + (-1.5) \hat{j} = -1.5\hat{j}\text{ ms}^{-1} \\ \text{and} \quad & \vec{r} = x \hat{i} + y \hat{j} = (4.5\hat{i} - 2.25\hat{j})\text{ m} \end{aligned}$
Hence, (A) and (D) are correct.

C,D

Since the particles have same ranges.
$\begin{array}{r l} & {\mathrm{Hence} \alpha + \beta = \frac {\pi}{2}} \\ & {\qquad R = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g} = \frac {2 u ^ {2} \sin \beta \cos \beta}{g}} \\ & {\qquad h _ {1} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}} \\ & {\qquad h _ {2} = \frac {u ^ {2} \sin^ {2} \beta}{2 g} = \frac {u ^ {2} \cos^ {2} \alpha}{2 g}} \\ {\Rightarrow} & {\quad R = 4 \sqrt {h _ {1} h _ {2}}} \\ & {\qquad t _ {1} = \frac {2 u \sin \alpha}{g}} \\ & {\qquad t _ {2} = \frac {2 u \sin \beta}{g} = \frac {2 u \cos \alpha}{g}} \\ {\Rightarrow} & {\quad \frac {t _ {1}}{t _ {2}} = \tan \alpha \text {and} \frac {h _ {1}}{h _ {2}} = \tan^ {2} \alpha} \end{array}$
$\Rightarrow \tan \alpha = \sqrt {\frac {h _ {1}}{h _ {2}}}$
Hence, (A), (B), (C) and (D) are correct.

$\Rightarrow \frac {x ^ {2}}{a ^ {2}} + \frac {y ^ {2}}{b ^ {2}} = 1$
$\ddot {y} = - p ^ {2} y$
so acceleration is directed towards the focus. Hence, (A), (B) and (C) are correct.

$0 = \sqrt {2 g h} (\cos \alpha) t - \frac {1}{2} (g \cos \alpha) t ^ {2}$
$\Rightarrow \quad t = \sqrt {\frac {8 h}{g}} \text { after it strikes the incline at A }$
$s = \sqrt {2 g h} (\sin \alpha) t + \frac {1}{2} (g \sin \alpha) t ^ {2}$
$\text { Substituting value of t,we get } s = 8 h \sin \alpha$
Hence,(A)and(D)are correct.