Kinematics-1D

A,D

A,B,D

A,B,C,D

B,C,D

B,C,D

A,B,C

Ans. (C)

$15 \times t_{0} = 30 \Rightarrow t_{0} = 2 \sec$
and
$\begin{array}{l} \mathrm {v_ {y}} = 1 0 \mathrm{m/sec} \\ \mathrm {v_ {x}} = 1 \mathrm{s} \end{array}$
$\Rightarrow \tan \theta = \frac {2}{3}$
$t _ {0} = \frac {2 v _ {y}}{g}$

Ans. (C)

$t _ {0} = \frac {2 v _ {y}}{g}$
$\Rightarrow \quad v _ {y} = 1 0 m / \sec$
$v _ {x} = 1 5 \text { and } H _ {\max} = 7 5 + \frac {(1 0) ^ {2}}{2 g} = 8 0$
$T = 2 \sqrt {\frac {2 H _ {m a x}}{g}} = 2 \sqrt {\frac {8 0 \times 2}{1 0}} = 8 s e c$
$A B = v _ {x} \times T = 8 \times 1 5 = 1 2 0 m$

Ans. (B)

$v _ {x \mathrm{rel}} = (v _ {x}) _ {\mathrm{stone}} - v _ {\mathrm{car}} = 0$
$\Rightarrow$ Car is just above the stone
$\Rightarrow \quad \text { Height } = 7 5 m$

Ans. (B)
Range = 200
$\frac {u ^ {2} \sin 2 \theta}{g} = 2 0 0$
$\sin 2 \theta = \frac {2 0 0}{2 5 0}$
h = 2 5 0

$\frac {u ^ {2}}{g} = 2 5 0$
$2 \theta = 5 3 ^ {\circ}$
$\theta_ {2} = 9 0 - \theta = 9 0 - 2 6. 5$
$\theta_ {2} = 6 3. 5$

Ans. (D)
$T = \frac {2 u \sin \theta}{g}$
$T \propto \sin \theta$

MISCONCEPTION
Taking east direction as positive, we have
$u = + 9 \mathrm{ms} ^ {- 1} \text { and } a = - 3 \mathrm{ms} ^ {- 2}$
$s _ {5 ^ {t h}} = u + \frac {\alpha}{2} (2 (5) - 1)$
$\Rightarrow s _ {5 ^ {t h}} = 9 - \frac {2}{2} (9) = 0$
Actually this is not the correct answer, because we must observe that the particle reverses its direction of motion (v=0) at t=4.5 s which lies in the specified interval. So, we will proceed other way round.
Since the total distance is
d s = v d t
$\Rightarrow \quad s _ {\mathrm{total}} = \int v d t$
$\Rightarrow \quad s _ {\mathrm{total}} = \left| \int_ {4} ^ {4 \cdot 5} v d t \right| + \left| \int_ {4 \cdot 5} ^ {5} v d t \right|$
$\mathrm{Now} \int v d t = \int (u + a t) d t = u t + \frac {1}{2} a t ^ {2}$
$\begin{array}{r l r} & {\Rightarrow} & {s _ {\mathrm{total}} = \bigg | 9 (4. 5 - 4) + \frac {1}{2} (- 2) \bigg | \bigg [ \bigg (\frac {9}{2} \bigg) ^ {2} - (4) ^ {2} \bigg ] +} \\ & & {\qquad \bigg | 9 (5 - 4. 5) + \frac {1}{2} (- 2) \bigg [ (5) ^ {2} - \bigg (\frac {9}{2} \bigg) ^ {2} \bigg ] \bigg |} \\ & {\Rightarrow} & {s _ {\mathrm{total}} = | 4. 5 - 4. 2 5 | + | 4. 5 - 4. 2 5 | = 0. 5 \mathrm{m}} \end{array}$
Hence, the correct answer is (B).

$\begin{array}{r l} & {a = \frac {d v}{d t} = 8 - 2 t} \\ & {\Rightarrow a = 8 - 2 (5)} \\ & {\Rightarrow a = - 2 \mathrm{ms} ^ {- 2}} \end{array}$
Hence, the correct answer is (D).

For constant speed, $a_{T} = 0$
$\begin{array}{r l} \Rightarrow & a = a _ {C} = a _ {N} = \frac {V _ {T} ^ {2}}{R} \\ \Rightarrow & 0. 8 g = \frac {v ^ {2}}{(2 0 0)} \\ \Rightarrow & v ^ {2} = (0. 8) (1 0) (2 0 0) \\ \Rightarrow & v = 4 0 \mathrm{ms} ^ {- 1} \\ \Rightarrow & v = 4 0 \times \frac {1 8}{5} \mathrm{kmh} ^ {- 1} \\ \Rightarrow & v = 1 4 4 \mathrm{kmh} ^ {- 1} \end{array}$
Hence, the correct answer is (C).

Let the balls A and B collide at the point P a distance h from the ground. Let u be the velocity of launch of B and the launch velocity of A is zero. At the point P, final velocity of A is
$v _ {A} = \sqrt {2 g (H - h)}\tag{... (1}$

and that of $B$ is
$v _ {B} = \sqrt {u ^ {2} - 2 g h}\tag{... (2}$
Also, for the balls to collide at P, time taken by A to reach P equals time taken by B to reach P. So we have
$(H - h) = \frac {1}{2} g t ^ {2}\tag{... (3}$
$h = u t - \frac {1}{2} g t ^ {2}\tag{... (4}$
$(3), t = \sqrt {\frac {2 (H - h)}{g}}$
From
Substitute in (4), we get
$\begin{array}{r l} & h = u \sqrt {\frac {2 (H - h)}{g}} - \frac {1}{2} g \frac {2 (H - h)}{g} \\ \Rightarrow & h = u \sqrt {\frac {2 (H - h)}{g}} - (H - h) \\ \Rightarrow & h = u \sqrt {\frac {2 (H - h)}{g}} - H + h \\ \Rightarrow & u = H \sqrt {\frac {g}{2 (H - h)}} \end{array}\tag{... (5}$
Substitute (5) in (2), we get
$v _ {B} = \sqrt {\frac {g H ^ {2}}{2 (H - h)} - 2 g h}\tag{... (6}$
$\begin{array}{r l} & {\mathrm{Since} v _ {A} = 2 v _ {B}} \\ & {\Rightarrow v _ {A} ^ {2} = 4 v _ {B} ^ {2}} \\ & {\Rightarrow 2 g (H - h) = \frac {4 H ^ {2} g}{2 (H - h)} - 8 g h} \\ & {\Rightarrow 2 g H - 2 g h = \frac {2 H ^ {2} g}{H - h} - 8 g h} \\ & {\Rightarrow H - h = \frac {H ^ {2}}{H - h} - 4 h} \\ & {\Rightarrow \frac {H ^ {2}}{H - h} - H = 3 h} \\ & {\Rightarrow \frac {H ^ {2} - H ^ {2} + H h}{H - h} = 3 h} \\ & {\Rightarrow \frac {H}{H - h} = 3} \\ & {\Rightarrow H = 3 H - 3 h} \\ & {\Rightarrow 2 H = 3 h} \\ & {\Rightarrow h = \frac {2 H}{3}} \end{array}$
Hence, the correct answer is (D).

$\tan\beta=\frac{v_{wind}\sin\theta}{v_{boat}+v_{wind}\cos\theta}$

$\Rightarrow \quad \tan \beta = \frac {7 2 \sin (1 3 5 ^ {\circ})}{5 1 + 7 2 \cos (1 3 5 ^ {\circ})}$
$\Rightarrow \tan \beta \rightarrow \infty$
$\Rightarrow \quad \beta \approx 9 0 ^ {\circ}$
i.e., direction of flag is towards east
Hence, the correct answer is (A).

$L = \frac{1}{2} gt^2 -\frac{1}{2} g(t - T)^2$
Solve $t=\frac{T}{2}+\frac{L}{gT}$
Hence, the correct answer is (B).

Let the point $B$ be situated at a distance $x$ below $A$ . If the particle reaches $B$ in time $t$ , then
$x = \frac {1}{2} g t ^ {2}\tag{... (1}$

For AC, $x + h = \frac{1}{2} g(t + 2)^{2}$
$\ldots(2)$
For AD, $x + 2h = \frac{1}{2} g(t + 3)^{2}$
... (3)
Solving (1), (2) and (3), we get $t = 0.5\mathrm{s}$
Hence, the correct answer is (B).

At the maximum height, v = 0
$\Rightarrow 0 ^ {2} - u ^ {2} = 2 (- g) (h)$
$\begin{array}{r l} \Rightarrow & u = \sqrt {2 g h} \\ \Rightarrow & u = \sqrt {2 (1 0) (2 0)} = 2 0 \mathrm{ms} ^ {- 1} \end{array}$
If t be the time taken by the ball to reach the maximum height, then
0 = u - g t
$\Rightarrow 0 = 2 0 - (1 0) t$
$\Rightarrow t = 2 \mathrm{s}$
So, time taken by the ball to return to juggler's hand is 4 s. To maintain a proper distance between them, the balls must be thrown up at an interval of
$\Delta t = \frac {4}{4} s = 1 \mathrm{s}$
Hence, the correct answer is (B).

C