Kinematics-2D

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.

$\vec {u} = 4 \hat {i} + 3 \hat {j} + 3 \sqrt {3} \hat {k}$
$\Rightarrow \quad | \vec {u} | = 2 \sqrt {1 3} \mathrm{ms} ^ {- 1}$
$\text {Velocity at highest point is} \vec {v} = 4 \hat {i} + 3 \hat {j}$
$\{\text {because at the highest point v_{Z}} = 0 \}$
$\Rightarrow \quad | \vec {v} | = 5 \mathrm{ms} ^ {- 1}$
Hence, (B) and (D) are correct.

Let shells collide in mid-air at point $P(x, y)$. For the 1st shell:
$x = (u \cos \alpha) T \tag{1}$
$y = (u \sin \alpha) T - \frac{1}{2} g T^2 \tag{2}$
For the 2nd shell:
$x = u \cos \beta (T - t) \tag{3}$
$y = (u \sin \beta) (T - t) - \frac{1}{2} g (T - t)^2 \tag{4}$
Equating (1) and (3), we get:
$T \cos \alpha = (T - t) \cos \beta \tag{\text{OPTION (A)}}$
$\begin{aligned} \Rightarrow \frac{\cos \alpha}{\cos \beta} &= 1 - \frac{t}{T} \\ \Rightarrow \frac{\cos \alpha}{\cos \beta} &< 1 \\ \Rightarrow \cos \alpha &< \cos \beta \\ \Rightarrow \alpha &> \beta \end{aligned} \tag{\text{OPTION (B)}}$
Equating (2) and (4), we get:
$(u \sin \alpha) T - \frac{1}{2} g T^2 = (u \sin \beta) (T - t) - \frac{1}{2} g (T - t)^2 \tag{\text{OPTION (D)}}$
Hence, (A), (B) and (D) are correct.

A,B,C,D

Let the projectile reach the point of impact in time t. Then
$\begin{array}{r l r} & & {l \sin \theta = \frac {1}{2} g t ^ {2}} \\ & {\Rightarrow} & {t = \sqrt {\frac {2 \ell \sin \theta}{g}}} \end{array}\tag{... (1}$

Further
$\begin{array}{r l} & l \cos \theta + D = v t \\ \Rightarrow & D = v t - l \cos \theta \\ \Rightarrow & D = v \sqrt {\frac {2 l \sin \theta}{g}} - l \cos \theta \end{array}\tag{... (2}$
For projectile to pass over the observer's head
$\begin{array}{r l} & D > 0 \\ \Rightarrow & v \sqrt {\frac {2 l \sin \theta}{g}} > l \cos \theta \end{array}$
$\Rightarrow v ^ {2} \left(\frac {2 l \sin \theta}{g}\right) > l ^ {2} \cos^ {2} \theta$
\Rightarrow l < \frac {2 v ^ {2} \tan \theta \sec \theta}{g}
Hence, (B) and (C) are correct.

Since, we know that the range is same for complimentary angles i.e., $R_{\theta} = R_{90-\theta} = \frac{2u^{2}\sin\theta\cos\theta}{g}$
Then, $T_{1} = \frac{2u\sin\theta}{g}$ and $T_{2} = \frac{2u\cos\theta}{g}$
$\Rightarrow T_{1}T_{2}\propto R$ and
$\frac {T _ {1}}{T _ {2}} = \tan \theta$
Hence, (C) and (D) are correct.

Let the second shell reach the point $P(x, y)$ at time $t_{2}$ and the first shell be fired at a time
$\left(t _ {2} - \Delta t\right) = t _ {1}.$

For shell 1
$x = \left(5 \sqrt {3}\right) t_1 \tag{1}$
$-(10 - y) = - \frac{1}{2} g t_1^2 \tag{2}$
For shell 2
$x = \left(5 \sqrt{3} \cos 60^\circ\right) t_2 \tag{3}$
$-(10 - y) = (5 \sqrt{3} \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2 \tag{4}$
Equating (1) and (3), we get:
$t_2 = 2 t_1 \tag{5}$
Equating (2) and (4), we get:
$\begin{aligned} - \frac{1}{2} g t_1^2 &= (5 \sqrt{3} \sin 60^\circ) t_2 - \frac{1}{2} g t_2^2 \\ \Rightarrow - \frac{1}{2} g t_1^2 &= \frac{15}{2} (2 t_1) - \frac{1}{2} g (2 t_1)^2 \\ \Rightarrow - \frac{1}{2} g t_1^2 &= 15 t_1 - 2 g t_1^2 \\ \Rightarrow - 5 t_1^2 &= 15 t_1 - 20 t_1^2 \\ \Rightarrow 15 t_1^2 &= 15 t_1 \\ \Rightarrow 15 t_1^2 - 15 t_1 &= 0 \end{aligned}$
$\Rightarrow 15 t_1 (t_1 - 1) = 0$
$\Rightarrow t_1 = 1\text{ s}$
$\Rightarrow t_2 = 2\text{ s} \quad \text{i.e., } \Delta t = 1\text{ s}$
$\Rightarrow x = 5 \sqrt{3}\text{ m} \quad \text{and} \quad y = 5\text{ m}$
Hence, (A), (B), (C) and (D) are correct.

Since $R=\frac{u^2\sin(2\theta)}{g}$,
$\Rightarrow R=kg^{-1} \tag{1}$
$\Rightarrow dR=-kg^{-2},dg \tag{2}$
$\Rightarrow \frac{dR}{R}=\frac{-kg^{-2},dg}{kg^{-1}}$
$\Rightarrow \frac{dR}{R}=-\frac{dg}{g}
=-\left(\frac{g_p-g_e}{g}\right)$
$\Rightarrow \text{Fractional Decrement}=\frac{dR}{R}=\frac{1}{291}$
Range at equator is
$R_e=\frac{u^2\sin(2\theta)}{g_e}$
Range at pole is
$R_p=\frac{u^2\sin(2\theta)}{g_p}$
$R = \left(\begin{array}{c} \text{Original range with} \\ \text{no variation of }g \\ \text{taken into account} \end{array}\right) = \frac{u^2\sin(2\theta)}{g}$
$\frac{dR}{R} = \left(\begin{array}{c} \text{Fractional} \\ \text{Decrement} \end{array}\right) = \frac{R_e-R_p}{R}$
$\Rightarrow \frac{dR}{R} = \frac{u^2\sin(2\theta)\left(\frac{1}{g_e}-\frac{1}{g_p}\right)} {u^2\sin(2\theta)\frac{1}{g}}$
$\Rightarrow \frac{dR}{R} = -g\left(\frac{1}{g_p}-\frac{1}{g_e}\right)$
Hence, (B) and (D) are correct.

For shell to hit the boat the time lapsed equals the time of flight of the shell = $\frac{2v_{2}\sin\alpha}{g}$ $\Rightarrow$ Distance travelled by boat = $v_{1}\left(\frac{2v_{2}\sin\alpha}{g}\right)$ Distance of boat from the cannon at the instant the shell is fired is $x = R - \left(\frac{2v_{1}v_{2}\sin\alpha}{g}\right)$
$\begin{array}{l l} \Rightarrow & x = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha) - \frac {2 v _ {1} v _ {2} \sin \alpha}{g} \\ \Rightarrow & x = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha - v _ {1}) \end{array}$
and the distance of boat from the cannon when the shell hits the boat is equal to the range of the shell
$R = \frac {2}{g} (v _ {2} \sin \alpha) (v _ {2} \cos \alpha)$
Hence, (A), (B), (C) and (D) are correct.

$\begin{array}{r l} & u _ {x} = u \cos (4 5 ^ {\circ}) = 4 0 \mathrm{ms} ^ {- 1} \text {and} \\ & u _ {y} = u \sin (4 5 ^ {\circ}) = 4 0 \mathrm{ms} ^ {- 1} \\ & \text {At} t = 2 \mathrm{s}, \\ & v _ {x} = u _ {x} = 4 0 \mathrm{ms} ^ {- 1} \\ & v _ {y} = u _ {y} - g t = 4 0 - (1 0) (2) = 2 0 \mathrm{ms} ^ {- 1} \end{array}\tag{... (1}$
So, (B) is correct
$\mathrm{Further} x = u _ {x} t = 8 0 \mathrm{m}$
$\mathrm{and} y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2}$
$\Rightarrow y = (4 0) (2) + \frac {1}{2} (- 1 0) (4)$
$\Rightarrow y = 6 0 \mathrm{m}$
$\mathrm{Hence(D)iscorrect}$
So, displacement of the particle is $r = \sqrt{x^{2} + y^{2}}$
$\Rightarrow r = \sqrt {(8 0) ^ {2} + (6 0) ^ {2}} = 1 0 0 \mathrm{m}$
Hence (A) is also correct.
Also, we observe that if $\alpha$ is the angle made by v with the horizontal, then
$\tan \alpha = \frac {v _ {y}}{v _ {x}} = \frac {2 0}{4 0} = \frac {1}{2}$
Hence (C) is not correct.
Hence, (A), (B) and (D) are correct.

$x_0 + ut = vt$
$\begin{array}{l} x _ {0} + u t = v t \\ \Rightarrow \quad \left(6 \hat {i} + 8 \hat {j}\right) + (2 t) \hat {i} = \left(a \hat {i} + b \hat {j}\right) t \\ \text { At } t = 2 \text { s, we get } \\ \Rightarrow \quad 1 0 \hat {i} + 8 \hat {j} = 2 a \hat {i} + 2 b \hat {j} \\ \Rightarrow \quad 2 a = 1 0 \text { and } 2 b = 8 \\ \Rightarrow \quad a = 5 \text { and } b = 4 \\ \text { Hence, (B) and (C) are correct. } \end{array}$

$y = h - x\tan \alpha$
$\mathrm{Further} y = \big (v _ {0} \sin \theta \big) t - \frac {1}{2} g t ^ {2}\tag{... (2}$
and $x = (v_{0} \cos \theta)t$
Equating (1) and (2) and put value of x from (3) to get
$\Rightarrow t ^ {2} - \frac {2 v _ {0}}{g} (\cos \theta \tan \alpha + \sin \theta) t + \frac {2 h}{g} = 0$
$\Rightarrow t ^ {2} - \left[ \frac {2 v _ {0} \sin (\theta + \alpha)}{g \cos \alpha} \right] t + \frac {2 h}{g} = 0\tag{... (4}$
For $t$ to be MINIMUM
$\begin{array}{l l} & \frac {d t}{d \theta} = 0 \\ \Rightarrow & 2 t \frac {d t}{d \theta} - \frac {2 v _ {0}}{g \cos \alpha} \frac {d}{d \theta} [ t \sin (\alpha + \theta) ] + 0 = 0 \\ \Rightarrow & 0 - \frac {2 v _ {0}}{g \cos \alpha} \left[ \sin (\alpha + \theta) \frac {d t}{d \theta} + t \cos (\alpha + \theta) \right] = 0 \\ \Rightarrow & 0 - \frac {2 v _ {0} t \cos (\alpha + \theta)}{g \cos \alpha} = 0 \quad \left\{\because \frac {d t}{d \theta} = 0 \right\} \\ \Rightarrow & \cos (\theta + \alpha) = 0 \quad \{\because t \neq 0 \} \\ \Rightarrow & \theta + \alpha = \frac {\pi}{2} \quad \{\text {OPTION (A)} \} \end{array}$
Substituting $\theta +\alpha = \frac{\pi}{2}$ in (4), we get
$t _ {\mathrm{min}} ^ {2} - \frac {2 v _ {0}}{g \cos \alpha} t _ {\mathrm{min}} + \frac {2 h}{g} = 0$
$\Rightarrow t _ {\mathrm{min}} = \frac {v _ {0} - \sqrt {v _ {0} ^ {2} - 2 g h \cos^ {2} \alpha}}{g \cos \alpha}$
{OPTION (C)}
For $t_{\mathrm{min}}$ to be real $v_0^2 > 2gh\cos^2\alpha$
$\Rightarrow v _ {0} > \sqrt {2 g h} \cos \alpha$
So if $v_{0} < \sqrt{2gh}\cos\alpha$ , then projectile never reaches the roof.
{OPTION (D)}
Hence, (A), (C) and (D) are correct.

As studied and discussed in theory, we observe that all (A), (B), (C) and (D) are correct.
Hence, (A), (B), (C) and (D) are correct.

The vertical velocity at P is zero.
Hence $v \sin (\theta + \alpha) - gt = 0$
$\Rightarrow t = \frac {v \sin (\theta + \alpha)}{g}\tag{... (1}$

Since displacement perpendicular to the plane is zero hence
$0 = (v \sin \theta) t + \frac {1}{2} (- g \cos \alpha) t ^ {2}$
$\Rightarrow t = \frac {2 v \sin \theta}{g \cos \alpha}\tag{... (2}$
Equating (1) & (2) and rearranging, we get
$\frac {v \sin (\theta + \alpha)}{g} = \frac {2 v \sin \theta}{g \cos \alpha}$
$\Rightarrow 2 \sin \theta = \cos \alpha (\sin \theta \cos \alpha + \cos \theta \sin \alpha)$
$\Rightarrow 2\sin \theta = \cos^2\alpha \sin \theta +\cos \theta \sin \alpha \cos \alpha$
$\Rightarrow \quad (2 - \cos^ {2} \alpha) \sin \theta = \cos \theta \sin \alpha \cos \alpha$
$\Rightarrow \tan \theta = \frac {\sin \alpha \cos \alpha}{2 - \cos^ {2} \alpha}$
$\Rightarrow \tan \theta = \frac {\sin \alpha \cos \alpha}{(1 - \cos^ {2} \alpha) + 1} = \frac {\sin \alpha \cos \alpha}{1 + \sin^ {2} \alpha}$
$\Rightarrow \quad \tan \theta = \frac {\tan \alpha}{1 + 2 \tan^ {2} \alpha}$
Hence, (A) and (D) are correct.

Time of ascent $= 2+1=3,s$
If $\alpha$ is the angle of projection, then
$\frac{u\sin\alpha}{g}=3$
$\Rightarrow u\sin\alpha=30 \tag{1}$
If $\beta=30^\circ$ is the angle which the projectile makes with the horizontal at $t=2,s$ and $v$ is the velocity at this instant of time, then
$v\cos\beta=u\cos\alpha \tag{2}$
and
$v\sin\beta=u\sin\alpha-g(2)$
$\Rightarrow v\sin\beta=30-20=10$
$\Rightarrow v\sin30^\circ=10$
$\Rightarrow v=20,ms^{-1}$
Substituting $\beta=30^\circ$ and $v=20,ms^{-1}$ in (2), we get
$v\frac{\sqrt{3}}{2}=u\cos\alpha$
$\Rightarrow u\cos\alpha=10\sqrt{3} \tag{3}$
Squaring (1) and (3) and adding,
$u^2=30^2+(10\sqrt{3})^2$
$\Rightarrow u^2=1200$
$\Rightarrow u=20\sqrt{3},ms^{-1}$
and
$\tan\alpha=\frac{u\sin\alpha}{u\cos\alpha}$
$=\frac{30}{10\sqrt{3}}$
$=\sqrt{3}$
$\Rightarrow \alpha=60^\circ$ with horizontal or $30^\circ$ with vertical.
Hence, (B) and (C) are correct.

At the instant shown, we have

$v \cos \theta = u \cos (6 0 ^ {\circ}) = \frac {u}{2}$
$\Rightarrow v = \frac {u}{2 \cos \theta}\tag{... (1}$
The radius of curvature is given by
$R = \frac {v ^ {2}}{g \cos \theta} = \frac {u ^ {2}}{4 g \cos^ {3} \theta}$
$\Rightarrow R _ {\min} = \frac {u ^ {2}}{4 g}, \text { when } \theta = 0 ^ {\circ}$
$\mathrm{Since}, R = \frac {8}{3 \sqrt {3}} R _ {\mathrm{min}}$
$\Rightarrow \frac {u ^ {2}}{4 g \cos^ {3} \theta} = \frac {8}{3 \sqrt {3}} \left(\frac {u ^ {2}}{4 g}\right)$
$\Rightarrow \cos^ {3} \theta = \frac {3 \sqrt {3}}{8} = \left(\frac {\sqrt {3}}{2}\right) ^ {3}$
$\Rightarrow \cos \theta = \frac {\sqrt {3}}{2}$
$\Rightarrow \theta = 3 0 ^ {\circ}$
So, from equation (1), we get
$v = \frac {u}{\sqrt {3}}$
Hence, (B) and (C) are correct.

$\Rightarrow \frac {d y}{d t} = \frac {3 d x}{d t}$
$\Rightarrow v _ {y} = 3 v _ {x} = 3$
$\left\{\because v _ {2} = 1 \mathrm{ms} ^ {- 1} \right\}$
Accelerated curvilinear
$\begin{array}{r l} \Rightarrow & \frac {d y _ {y}}{a t} = \frac {3 d v _ {x}}{d t} \\ \Rightarrow & a _ {y} = 3 a _ {x} = 3 \end{array}$
$\left\{\because a _ {x} = 1 \mathrm{ms} ^ {- 1} \right\}$
nothing can be said about velocity and acceleration as the given values are at particular instant.