Kinematics-2D

Since $R=\frac{2u\sin(\alpha-\beta)\cos\alpha}{g\cos^{2}\beta}$ , where $u=2ms^{-1}$ $\alpha=30+15=45^{\circ}$ $\beta=30^{\circ}$ $\Rightarrow R=\frac{2(2)^{2}\sin(15^{\circ})\cos(45^{\circ})}{g\cos^{2}(30^{\circ})}$ $\Rightarrow R=\frac{8\sqrt{2}}{15}\sin(15^{\circ})$ $\Rightarrow R=\frac{8\sqrt{2}}{15}\sin(45^{\circ}-30^{\circ})$ $\Rightarrow R=\frac{8\sqrt{2}}{15}\left(\frac{1}{2\sqrt{2}}\right)(\sqrt{3}-1)=\frac{4}{15}(\sqrt{3}-1)$ $\Rightarrow R\approx20cm$ Hence, the correct answer is (B).
Hence, the correct answer is (B).

For same horizontal range, we must have angles as complimentary. So
For $\theta_{1} = \theta$ and $\theta_{2} = (90 - \theta)$
$\begin{array}{l l} & R = \frac {2 u ^ {2} \sin \theta \cos \theta}{g} \\ \Rightarrow & t _ {1} = \frac {2 u \sin \theta}{g} \text { and } t _ {2} = \frac {2 u \cos \theta}{g} \\ \Rightarrow & t _ {1} t _ {2} = \frac {u ^ {2} 4 \sin \theta \cos \theta}{g ^ {2}} \\ \Rightarrow & t _ {1} t _ {2} = \frac {2 u ^ {2} \sin 2 \theta}{g ^ {2}} = \frac {2 R}{g} \end{array}$
Hence, the correct answer is (B).

$y = 2x - 9x^{2}$
Comparing it with standard equation of trajectory of projectile, i.e.,
$y = x \tan \theta - \frac {g x ^ {2}}{2 4 ^ {2} \cos^ {2} \theta}$

we get, $\tan \theta = 2$
and $9=\frac{10\times5}{2v_{0}^{2}}$
$\Rightarrow v _ {0} = \frac {5}{3} \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (C).

For complementary angles, ranges are equal.
So, for $\theta_{1} = \theta$ and $\theta_{2} = (90 - \theta)$ , we have
$R = \frac {2 u ^ {2} \sin \theta \cos \theta}{g}$
$h _ {1} = \frac {u ^ {2} \sin^ {2} \theta}{2 g}, h _ {2} = \frac {u ^ {2} \cos^ {2} \theta}{2 g}$
$\Rightarrow h _ {1} h _ {2} = \left(\frac {2 u ^ {2} \sin \theta \cos \theta}{g}\right) ^ {2} \times \left(\frac {1}{1 6}\right)$
$\Rightarrow 1 6 h _ {1} h _ {2} = R ^ {2}$
$\Rightarrow R ^ {2} = 1 6 \left(h _ {1} h _ {2}\right)$
Hence, the correct answer is (B).

Maximum areas will be covered when range is maximum, so
$\begin{array}{r l} & R _ {\max} = \frac {u ^ {2}}{g} \\ \Rightarrow & \frac {A _ {1}}{A _ {2}} = \frac {\pi (R _ {\max}) _ {1} ^ {2}}{\pi (R _ {\max}) _ {2} ^ {2}} = \left(\frac {u _ {1} ^ {2}}{u _ {2} ^ {2}}\right) ^ {2} = \frac {u _ {1} ^ {4}}{u _ {2} ^ {4}} = \frac {1}{1 6} \end{array}$
Hence, the correct answer is (D).

Speed $v = 10 \, ms^{-1}$
Since, centripetal acceleration is given by,
$a = \frac {v ^ {2}}{r}$

and $|\vec{v}| = \mathrm{constant}$
$\Rightarrow a \propto \frac {1}{r}$

Since, $\vec{u} = \hat{i} + 2\hat{j}$
$\mathrm{Also}, \vec {u} = u _ {x} \hat {i} + u _ {y} \hat {j}$
$\Rightarrow u _ {x} = 1 \text { and } u _ {y} = 2$
$\text { Now } x = u _ {x} t \text { and } y = u _ {y} t - \frac {1}{2} g t ^ {2}$
and $y = 2t - \frac{1}{2} \times 10 \times t^{2} = 2t - 5t^{2}$
$\Rightarrow \quad y = 2 x - 5 x ^ {2}$
So, equation of trajectory is $y = 2x - 5x^{2}$
Hence, the correct answer is (C).

Let u be the velocity of projection of the stone. The maximum height a boy can throw a stone is
$H _ {\mathrm{max}} = \frac {u ^ {2}}{2 g} = 1 0 \mathrm{m}\tag{... (1}$
The maximum horizontal distance the boy can throw the same stone is
$R _ {\mathrm{max}} = \frac {u ^ {2}}{g} = 2 0 \mathrm{m}\tag{{Using (1)}}$
Hence, the correct answer is (C).

$R _ {\max} = \frac {v ^ {2} \sin 9 0 ^ {\circ}}{g} = \frac {v ^ {2}}{g}$
$\mathrm{Area} = \pi \big (R _ {\max} \big) ^ {2} = \frac {\pi v ^ {4}}{g ^ {2}}$
Hence, the correct answer is (B).

The position vector of the particle from the origin at any time t is
$\begin{array}{r l} & {\vec {r} = v _ {0} t \cos \theta \hat {i} + \left(v _ {0} t \sin \theta - \frac {1}{2} g t ^ {2}\right) \hat {j}} \\ & {\mathrm{Velocityvector}, \vec {v} = \frac {d \vec {r}}{d t}} \\ & {\Rightarrow \quad \vec {v} = \frac {d}{d t} \bigg (v _ {0} t \cos \theta \hat {i} + \left(v _ {0} t \sin \theta - \frac {1}{2} g t ^ {2}\right) \hat {j} \bigg)} \\ & {\Rightarrow \quad \vec {v} = v _ {0} \cos \theta \hat {i} + (v _ {0} \sin \theta - g t) \hat {j}} \end{array}$
The angular momentum of the particle about the origin is
$\begin{array}{r l} & {\vec {L} = \vec {r} \times m \vec {v}} \\ {\Rightarrow} & {\vec {L} = m (\vec {r} \times \vec {v})} \\ {\Rightarrow} & {\vec {L} = m \bigg [ \bigg (v _ {0} t \cos \theta \hat {i} + \bigg (v _ {0} t \sin \theta - \frac {1}{2} g t ^ {2} \bigg) \hat {j} \bigg) \times} \\ & {\qquad \qquad \qquad \qquad \qquad \left(v _ {0} \cos \theta \hat {i} + (v _ {0} \sin \theta - g t) \hat {j}\right) \bigg ]} \\ {\Rightarrow} & {\vec {L} = m \Big [ (v _ {0} ^ {2} t \cos \theta \sin \theta - v _ {0} g t ^ {2} \cos \theta) \hat {k} +} \\ & {\qquad \qquad \qquad \qquad \qquad \left(v _ {0} ^ {2} t \sin \theta \cos \theta - \frac {1}{2} g t ^ {2} v _ {0} \cos \theta\right) (- \hat {k}) \bigg ]} \\ {\Rightarrow} & {\vec {L} = m \Big [ v _ {0} ^ {2} t \sin \theta \cos \theta \hat {k} - v _ {0} g t ^ {2} \cos \theta \hat {k} -} \\ & {\qquad \qquad \qquad \qquad v _ {0} ^ {2} t \sin \theta \cos \theta \hat {k} + \frac {1}{2} v _ {0} g t ^ {2} \cos \theta \hat {k} \Big ]} \\ {\Rightarrow} & {\vec {L} = m \Big [ - \frac {1}{2} v _ {0} g t ^ {2} \cos \theta \hat {k} \Big ] = - \frac {1}{2} m g v _ {0} t ^ {2} \cos \theta \hat {k}} \end{array}$
Hence, the correct answer is (D).

For a particle in uniform circular motion, acceleration is
$a = \frac {v ^ {2}}{R} (\text { towards the centre })$

From figure, we get
$\vec {a} = - a \cos \theta \hat {i} - a \sin \theta \hat {j} = - \frac {v ^ {2}}{R} \cos \theta \hat {i} - \frac {v ^ {2}}{R} \sin \theta \hat {j}$
Hence, the correct answer is (D).

$s=t^3+3$
$\Rightarrow v=\frac{ds}{dt}=\frac{d}{dt}(t^3+3)=3t^2$
Tangential acceleration,
$a_t=\frac{dv}{dt}=\frac{d}{dt}(3t^2)=6t$
At $t=2,\mathrm{s}$, we have
$v=3(2)^2=12,\mathrm{ms^{-1}}$
and
$a_t=6(2)=12,\mathrm{ms^{-2}}$
Also, centripetal acceleration is given by
$a_c=\frac{v^2}{R}=\frac{(12)^2}{20}=\frac{144}{20}=7.2,\mathrm{ms^{-2}}$
Net acceleration,
$a=\sqrt{(a_t)^2+(a_c)^2}=\sqrt{(7.2)^2+(12)^2}=14,\mathrm{ms^{-2}}$
Hence, the correct answer is (A).

$r = \frac {5}{2} \mathrm{m} = 2. 5 \mathrm{m}$
$a = r \omega^ {2} = (2. 5) \left(2 0 0 0 \times \frac {2 \pi}{6 0}\right) ^ {2}$
$\Rightarrow a = (2. 5) \frac {1 6 \pi^ {2}}{3 6} (1 0 0 0 0)$
$\Rightarrow a = 1. 1 \times 1 0 ^ {5} \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (B).

$h^{2}=r^{2}-x^{2}$
$\begin{array}{r l} & \tan \theta = \frac {h}{x} \\ & h = r \sin \theta \\ \Rightarrow & x = h \cot \theta \\ \Rightarrow & \frac {d x}{d t} = - h \sec^ {2} \theta \left(\frac {d \theta}{d t}\right) \\ \Rightarrow & \frac {d x}{d t} = - (8 0 0 0) \sec^ {2} (6 0) (- 0. 0 2 5) \\ \Rightarrow & \frac {d x}{d t} = (8 0 0 0) \left(\frac {4}{3}\right) (0. 0 2 5) \\ \Rightarrow & \frac {d x}{d t} = \frac {8 0 0}{3} \mathrm{ms} ^ {- 1} \\ \Rightarrow & \frac {d x}{d t} = \frac {8 0 0}{3} \times \frac {1 8}{5} \\ \Rightarrow & v = 9 6 0 \mathrm{kmh} ^ {- 1} \end{array}$
Hence, the correct answer is (B).

$\begin{array}{l} \omega = \frac {\left(v _ {\perp}\right) _ {r e l}}{r _ {\perp}} = \left[ \frac {\sqrt {3} \sin (6 0 ^ {\circ}) - 8 \sin (3 0 ^ {\circ})}{2 . 5} \right] \\ \Rightarrow \quad \omega = 1 \text { rads } ^ {- 1} \\ \text { Hence, the correct answer is (D). } \end{array}$

$Since 0 = (v \sin \theta) t + \frac {1}{2} (- a) t ^ {2}$
$\Rightarrow \quad t = \frac {2 v \sin \theta}{a}$
$\text {Also,} h = (v \cos \theta) t + \frac {1}{2} g t ^ {2}$
$\Rightarrow \quad h = \frac {2 v ^ {2}}{a} \sin \theta \left(\cos \theta + \frac {g}{a} \sin \theta\right)$
Hence, the correct answer is (D).

$\begin{array}{l} v _ {H} = \sqrt {\frac {2}{5}} \left(v _ {\frac {H}{2}}\right) \\ \text {Since speed at a point at height h is} \\ v = \sqrt {u ^ {2} - 2 g h} \\ \text {Since H} = \frac {u ^ {2} \sin^ {2} \theta}{2 g} \\ \Rightarrow v _ {H} = \sqrt {u ^ {2} - 2 g h} = u \cos \theta \text {and} \\ v _ {\frac {H}{2}} = \sqrt {u ^ {2} - 2 g \left(\frac {H}{2}\right)} = \sqrt {u ^ {2} - g H} \\ \Rightarrow v _ {\frac {H}{2}} = \sqrt {u ^ {2} - \frac {u ^ {2} \sin^ {2} \theta}{2}} \end{array}\tag{... (1}$
From (1)
$v _ {H} ^ {2} = \frac {2}{5} (v) _ {\frac {H}{2}} ^ {2}$
$\begin{array}{r l} \Rightarrow & u ^ {2} \cos^ {2} \theta = \frac {2}{5} \left(u ^ {2} - \frac {u ^ {2} \sin^ {2} \theta}{2}\right) \\ \Rightarrow & 5 \cos^ {2} \theta = 2 - \sin^ {2} \theta \\ \Rightarrow & 5 - 5 \sin^ {2} \theta = 2 - \sin^ {2} \theta \\ \Rightarrow & 4 \sin^ {2} \theta = 3 \\ \Rightarrow & \sin \theta = \frac {\sqrt {3}}{2} \\ \Rightarrow & \theta = 6 0 ^ {\circ} \end{array}$
Hence, the correct answer is (D).

If $T$ be the time of flight of the ball, then
$\begin{array}{r l} & T = \frac {2 v _ {y}}{g} = \frac {2 (v _ {0} \cos \theta)}{g} \\ \Rightarrow & T = \frac {2 (2 0 \cos \theta)}{1 0} = 4 \cos \theta \end{array}\tag{... (1}$
For the ball to return to the boy's hand, we have x = 0.
$\Rightarrow 0 = v _ {x} T + \frac {1}{2} a _ {x} T ^ {2}$

$\begin{array}{l l} \Rightarrow & 0 = (v _ {0} \sin \theta) T + \frac {1}{2} (- 4) T ^ {2} \\ \Rightarrow & 0 = 2 0 \sin \theta - 2 T \\ \Rightarrow & T = 1 0 \sin \theta \\ \text {From (1) and (2), we get} \\ & 4 \cos \theta = 1 0 \sin \theta \\ \Rightarrow & \cot \theta = 2. 5 \\ \Rightarrow & \theta = \cot^ {- 1} (2. 5) \\ \text {Hence, the correct answer is (D).} \end{array}\tag{... (2}$

Given $a_{N}=a_{T}$
$\begin{array}{l l} \Rightarrow & \frac {v ^ {2}}{R} = \frac {d v}{d t} \\ \Rightarrow & v ^ {- 2} d v = \frac {d t}{R} \\ \Rightarrow & \int_ {v _ {0}} ^ {v} v ^ {- 2} d v = \frac {1}{R} \int_ {0} ^ {t} d t \end{array}$
$\Rightarrow \frac {v ^ {- 2 + 1}}{- 2 + 1} \bigg | _ {v _ {0}} ^ {v} = \frac {1}{R} t \bigg | _ {0} ^ {t}$
$\Rightarrow \frac {1}{v _ {0}} - \frac {1}{v} = \frac {t}{R}$
$\Rightarrow v = \frac {v _ {0} R}{R - v _ {0} t}$
$\Rightarrow \frac {d r}{d t} = \frac {v _ {0} R}{R - v _ {0} t}$
$\Rightarrow \int_ {0} ^ {2 \pi R} d r = v _ {0} R \int_ {0} ^ {t} \frac {d t}{R - v _ {0} t}$
$\Rightarrow 2 \pi R = v _ {0} R \Bigg [ - \frac {1}{v _ {0}} \mathrm{log} _ {e} (R - v _ {0} t) \Bigg | _ {0} ^ {t} \Bigg ]$
$\Rightarrow 2 \pi R = - R \log_ {e} \left(\frac {R - v _ {0} t}{R}\right)$
$\Rightarrow - 2 \pi = \log_ {e} \left(\frac {R - v _ {0} t}{R}\right)$
$\Rightarrow \frac {R - v _ {0} t}{R} = e ^ {- 2 \pi}$
$\Rightarrow t = \frac {R}{v _ {0}} \big (1 - e ^ {- 2 \pi} \big)$
Hence, the correct answer is (D).

$R_{max} = \frac{v^{2}}{g}$ at $\theta = 45^{\circ}$
Maximum Area = $\pi(R_{\max})^2 = \frac{\pi v^4}{g^2}$
Hence, the correct answer is (B).

$\begin{array}{l} \vec {v} _ {a v} = \frac {\Delta \vec {x}}{\Delta t} = \frac {\text { Displacement }}{\text { Time lapsed }} \\ \Rightarrow | \vec {v} _ {a v} | = \frac {| \text { Displacement } |}{\text { Time lapsed }} = \frac {\sqrt {2} R}{t} \end{array}$

where $t = \sqrt{\frac{2\theta}{\alpha}}$
$\left\{\because \theta = \frac {1}{2} \alpha t ^ {2} \right\}$
Since $\theta = \frac{\pi}{2}$ radian and $\alpha = \frac{\pi}{4}$ rads $^{-2}$
$\Rightarrow t = \sqrt {\frac {2 \left(\frac {\pi}{2}\right)}{\left(\frac {\pi}{4}\right)}} = 2 \mathrm{s}$
$\Rightarrow \quad | \vec {v} _ {a v} | = \frac {\sqrt {2} R}{2} = \frac {\sqrt {2} \sqrt {2}}{2} = 1 \mathrm{ms} ^ {- 1}$
Hence, the correct answer is (A).

$R_{\phi} = R_{90 - \phi} = 4\sqrt{h_{\phi}h_{90 - \phi}}$
Hence, the correct answer is (A).