Kinematics-2D

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

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

Considering the vertical motion, if t be the time taken by the particle to go from A to B, then
$h = \frac {1}{2} g t ^ {2}$
$\Rightarrow t = \sqrt {\frac {2 h}{g}}$
During this time the particle must make an integral number of rotations (say $n$ ).
Now, considering the horizontal motion, then
$\begin{array}{r l r} & & u t = (2 \pi r) n \\ & \Rightarrow & n = \frac {u}{2 \pi r} \sqrt {\frac {2 h}{g}} \end{array}$
Hence, the correct answer is (A).

$\left\{\because \alpha = 4 5 ^ {o} \right\}$
$\begin{array}{l} R = (a + b) = \frac {u ^ {2} \sin (2 \alpha)}{g} = \frac {u ^ {2}}{g} \\ \Rightarrow u ^ {2} = g (a + b) \\ \text {Since,} y = a \tan \alpha - \frac {g a ^ {2}}{2 u ^ {2} \cos^ {2} \alpha} \\ \Rightarrow h = a \tan \alpha - \frac {g a ^ {2}}{2 g (a + b) \cos^ {2} \alpha} \\ \Rightarrow h = a \tan 4 5 - \frac {g a ^ {2}}{2 g (a + b) \cos^ {2} 4 5} \\ \Rightarrow h = \frac {a b}{a + b} \end{array}\tag{... (1}$
Hence, the correct answer is (D).

$\begin{array}{l} y = \frac {g x ^ {2}}{2 u ^ {2}} \\ \Rightarrow 2 0 - 1 2 = \frac {(9 . 8) (3 0) ^ {2}}{2 u ^ {2}} \\ \Rightarrow u ^ {2} = \frac {(9 . 8) 9 0 0}{8} \\ \Rightarrow u = 2 3. 5 \mathrm{ms} ^ {- 1} \\ \text { Hence,the correct answer } \end{array}$
Hence, the correct answer is (A).

Taking upward direction as $\oplus$ , we get
$\begin{array}{r l} & - h = v t + \frac {1}{2} (- g) t ^ {2} \\ \Rightarrow & t ^ {2} - \frac {2 v}{g} t - \frac {2 h}{g} = 0 \\ \Rightarrow & t = \frac {1}{2} \left(\frac {2 v}{g} + \sqrt {\frac {4 v ^ {2}}{g ^ {2}} + 4 \left(\frac {2 h}{g}\right)}\right) \\ \Rightarrow & t = \frac {1}{2} \left(\frac {2 v}{g} + \frac {2 v}{g} \sqrt {1 + \frac {2 h g}{v ^ {2}}}\right) \\ \Rightarrow & t = \frac {v}{g} \left(1 + \sqrt {1 + \frac {2 h g}{v ^ {2}}}\right) \end{array}$

Hence, the correct answer is (A).

At maximum height, radius of curvature is
$\begin{array}{l l} & R = \frac {v _ {T} ^ {2}}{a _ {C}} = \frac {(u \cos \theta) ^ {2}}{g} = \frac {u ^ {2} \cos^ {2} \theta}{g} = \frac {u _ {x} ^ {2}}{g} \\ \Rightarrow & R = \frac {\left(7 2 \times \frac {5}{1 8}\right) ^ {2}}{1 0} \\ \Rightarrow & R = \frac {4 0 0}{1 0} = 4 0 \mathrm{m} \end{array}$
Hence, the correct answer is (C).

Given $a_{y}=\frac{d^{2}y}{dt^{2}}=\alpha$ and $\frac{d^{2}x}{dt^{2}}=0$
Since $y = \beta x^{2}$
$\begin{array}{r l} \Rightarrow & \frac {d y}{d t} = 2 \beta x \left(\frac {d x}{d t}\right) \\ \Rightarrow & \frac {d ^ {2} y}{d t ^ {2}} = 2 \beta \left[ x \left(\frac {d ^ {2} x}{d t ^ {2}}\right) + \left(\frac {d x}{d t}\right) ^ {2} \right] \\ \Rightarrow & \alpha = 2 \beta v _ {x} ^ {2} \end{array}$
$\Rightarrow v _ {x} = \sqrt {\frac {\alpha}{2 \beta}}$
Hence, the correct answer is (B).

At maximum height, we have
$\begin{array}{r l} & u \cos \theta = \frac {u}{2} \\ \Rightarrow & \cos \theta = \frac {1}{2} \\ \Rightarrow & \theta = 6 0 ^ {\circ} \end{array}$
$R = \frac {\sqrt {3} u ^ {2}}{2 g}$
Hence, the correct answer is (C).