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).