Kinematics-2D

$\tan 30^{\circ} = \frac{u_{y}}{u_{x}}$
∴
$\frac {1}{\sqrt {3}} = \frac {8 0}{u _ {x}} \text {or} u _ {x} = 8 0 \sqrt {3} \mathrm{ms} ^ {- 1}$
$T = \frac {2 u _ {y}}{g} = \frac {2 \times 8 0}{1 0} = 1 6 \mathrm{s}$
$\mathrm{At} t = \frac {T}{4} = 4 \mathrm{s} \Rightarrow v _ {x} = u _ {x} = 8 0 \sqrt {3} \mathrm{ms} ^ {- 1} \text { and } v _ {y} = u _ {y} + a _ {y} t$
or $v_{y}=80+(-10)(4)=40\ ms^{-1}$
$\therefore \mathrm{Velocity} = \sqrt{(80\sqrt{3})^2 + (40)^2} = 144.2\mathrm{ms}^{-1}$

$(u \cos \alpha) = \sqrt{\frac{2}{5}} \sqrt{(u \cos \alpha)^{2} + \{(u \sin \alpha)^{2} - 2gh\}}$
... (i)
Here, $h=\frac{H}{2}=\frac{u^{2}\sin^{2}\alpha}{4g}$
... (ii)
Solving Eqs. (i) and (ii), we get
$\alpha = 6 0 ^ {\circ}$

The motion relative to elevator,
$a _ {r} = a _ {b} - a _ {e} = (- 1 0) - (+ 2) = - 1 2 \mathrm{ms} ^ {- 2}$
Now, $T=\frac{2u_{y}}{|a_{r}|}=\frac{2\times u\sin\theta}{|a_{r}|}=\frac{2\times4\times\sin30^{\circ}}{12}=\frac{1}{3}\mathrm{s}$

$R _ {\max} = \frac {u ^ {2}}{g} = 1. 6 \mathrm{m}$
(At $\theta = 45^{\circ}$ )
or $u = \sqrt{16} = 4\mathrm{ms}^{-1}$
Now,
$T = \frac {2 u _ {y}}{g} = \frac {2 (4 / \sqrt {2})}{1 0} = \frac {4}{5 \sqrt {2}} \mathrm{s}$
∴ Total distance travelled in $10 \, s = 1.6 \times \frac{10}{4/5\sqrt{2}} = 20 \, \sqrt{2} \, m$

For 5 s, weight of the body is balanced by the given force. Hence, it will move in a straight line as shown in the figure.

$\begin{array}{l} {R = \frac {u ^ {2} \sin 2 \theta}{g} + (u \cos \theta) (5)} \\ {= \frac {(5 0) ^ {2} \cdot \sin 6 0 ^ {\circ}}{1 0} + (5 0 \times \cos 3 0 ^ {\circ}) (5) = 2 5 0 \sqrt {3} \mathrm{m}} \end{array}$

From the given diagram, $\frac{R/2}{H}=\frac{\sqrt{3}H}{H}=\sqrt{3}$
or
$\frac {(v _ {0} ^ {2} \sin \theta \cos \theta) / g}{(v _ {0} ^ {2} \sin^ {2} \theta) / 2 g} = \sqrt {3} \Rightarrow 2 \cot \theta = \sqrt {3}$
or
$\tan \theta = \frac {2}{\sqrt {3}} \text {or} \theta = \tan^ {- 1} \left(\frac {2}{\sqrt {3}}\right)$

Given, $H = 2000 \, m$ , $u = 360 \, kmh^{-1}$

$u = 3 6 0 \times \frac {5}{1 8} = 1 0 0 \mathrm{ms} ^ {- 1}$
The time taken by the bomb to hit the target is
$T = \sqrt {\frac {2 H}{g}} = \sqrt {\frac {2 \times 2 0 0 0}{1 0}} = 2 0 \mathrm{s}$
The horizontal distance of the aeroplane from the target,
$R = u \sqrt {\frac {2 H}{g}} = 1 0 0 \times 2 0 = 2 0 0 0 \mathrm{m} = 2 \mathrm{km}$

Two balls will meet, if $(50 \cos 37^{\circ}) t_{A} = 120$ or $t_{A} = 3 s$
Vertical component of A is $50 \sin 37^{\circ}$ or $30 \, ms^{-1}$ , so they will meet, if thrown simultaneously.
$h _ {A} = h _ {B} = 3 0 \times 3 - \frac {1}{2} \times 1 0 \times (3) ^ {2} = 4 5 \mathrm{m}$

(a) Range becomes equal at complimentary angle. Hence,
$\beta = 9 0 ^ {\circ} - \alpha \text { or } \alpha + \beta = 9 0 ^ {\circ} = \pi / 2\tag{b}$
$h _ {1} = \frac {u ^ {2} \sin^ {2} \alpha}{2 g}$
$\Rightarrow h _ {2} = \frac {u ^ {2} \cos^ {2} \alpha}{2 g}$
$(\mathrm{As}, \beta = 9 0 ^ {\circ} - \alpha)$
$\therefore 4 \sqrt {h _ {1} h _ {2}} = \frac {2 u ^ {2} \sin \alpha \cos \alpha}{g} = \frac {u ^ {2} \sin 2 \alpha}{g} = R\tag{c}$
$\frac {t _ {1}}{t _ {2}} = \frac {(2 u \sin \alpha / g)}{(2 u \cos \alpha / g)} = \tan \alpha\tag{d}$
$\sqrt {\frac {h _ {1}}{h _ {2}}} = \tan \alpha$

Since, $\mathbf{v} \perp \mathbf{u}$ or $\mathbf{v} \cdot \mathbf{u} = 0$
or
$(\mathbf {u} + g t) \cdot \mathbf {u} = 0$
or
$\mathbf {u} \cdot \mathbf {u} + (g \cdot \mathbf {u}) t = 0$
or $u^{2} + gut \cos(90^{\circ} + \theta) = 0$
(Angle between u and g is $90^{\circ} + \theta$ )
or
$u - g t \sin \theta = 0$
:
$t = \frac {u}{g \sin \theta}$

Component (100 cos $53^{\circ}$ ) 60 ms $^{-1}$ will remain unchanged. Velocity will make $45^{\circ}$ with horizontal when vertical component also becomes $\pm 60$ ms $^{-1}$ .

Using $v = u + at$
(In vertical direction)
$+ 6 0 = 8 0 + (- 1 0) t _ {1}$
$\therefore t_{1} = 2 \mathrm{~s} \Rightarrow -60 = 80 + (-10) t_{2}$
$t _ {2} = 1 4 \mathrm{s}$

Area in which bullet will spread = $\pi r^{2}$
For maximum area, $r = R_{max} = \frac{v^{2}}{g}$ (When $\theta = 45^{\circ}$ )
Maximum area = $\pi R_{\max}^2 = \pi \left(\frac{v^2}{g}\right)^2 = \frac{\pi v^4}{g^2}$
$\begin{array}{l} \textbf {8 7 (d) A s ,} T = \frac {2 u _ {y}}{g} \Rightarrow u _ {y} = \frac {g T}{2} \\ \Rightarrow h _ {A} = u _ {y} t _ {A} - \frac {1}{2} g t _ {A} ^ {2} = u _ {y} \left(\frac {2 u _ {y}}{3 g}\right) - \frac {1}{2} g \left(\frac {2 u _ {y}}{3 g}\right) ^ {2} \\ = \frac {4}{9} \frac {u _ {y} ^ {2}}{g} = \left(\frac {4}{9 g}\right) \left(\frac {g T}{2}\right) ^ {2} = \frac {g T ^ {2}}{9} \quad \left(\because u _ {y} = \frac {g T}{2}\right) \\ \Rightarrow h _ {B} = u _ {y} \left(\frac {5}{6} \times \frac {2 u _ {y}}{g}\right) - \frac {1}{2} \times g \times \left(\frac {5}{6} \times \frac {2 u _ {y}}{g}\right) ^ {2} \\ = \frac {5}{1 8} \frac {u _ {y} ^ {2}}{g} = \frac {5}{1 8 g} \left(\frac {g T}{2}\right) ^ {2} = \frac {5}{7 2} g T ^ {2} \\ \therefore h _ {A} - h _ {B} = \frac {g T ^ {2}}{2 4} \end{array}$

D

The time taken by cart to cover 80 m,
$\frac {s}{v} = \frac {8 0}{3 0} = \frac {8}{3} \mathrm{s}$
The projectile must be fired (relative to cart) in vertically upward direction.
i.e. $a = -g = -10\mathrm{ms}^{-2},v^{\prime} = 0$ and $t = \frac{8 / 3}{2} = \frac{4}{3}\mathrm{s}$
$\therefore \quad v ^ {\prime} = u + a t \mathrm{or} 0 = u - 1 0 \times \frac {4}{3} \mathrm{or} u = \frac {4 0}{3} \mathrm{ms} ^ {- 1}$

$T/2 = 2 + 1 = 3 \, s$ or $T = 6 \, s$
$\Rightarrow \frac {2 u _ {y}}{g} = 6$
$\therefore \quad u _ {y} = 3 0 \mathrm{ms} ^ {- 1}$
Further, $\tan 30^{\circ} = \frac{v_y}{v_x} = \frac{u_y - gt}{u_x} = \frac{30 - 20}{u_x}$
or
or
$u = \sqrt {u _ {x} ^ {2} + u _ {y} ^ {2}} = 2 0 \sqrt {3} \mathrm{ms} ^ {- 1}$
$\tan \theta = \frac {u _ {y}}{u _ {x}} = \frac {3 0}{1 0 \sqrt {3}} = \sqrt {3} \text {or} \theta = 6 0 ^ {\circ}$

If the ball hits the nth step, the horizontal and vertical distances traversed are nb and nh, respectively. Let t be the time taken by the ball for these horizontal and vertical displacement. Then, velocity along horizontal direction remains constant = u, initial vertical velocity is zero.
$\therefore \qquad n b = u t$
$n h = 0 + (1 / 2) g t ^ {2}\tag{... (i}$
... (ii)
From Eqs. (i) and (ii), we get by eliminating t,
$n h = (1 / 2) g (n b / u) ^ {2} \Rightarrow n = \frac {2 h u ^ {2}}{g b ^ {2}}$

Given, speed of packets = 125 ms $^{-1}$
Height of the hill = 500 m.
To cross the hill, the vertical component of the velocity should be sufficient to cross such height.
$\begin{array}{r l} & u _ {y} \geq \sqrt {2 g h} \\ & \quad \geq \sqrt {2 \times 1 0 \times 5 0 0} \geq 1 0 0 \mathrm{ms} ^ {- 1} \end{array}$
But
$u ^ {2} = u _ {x} ^ {2} + u _ {y} ^ {2}$
∴ Horizontal component of initial velocity,
$\begin{array}{r l} & {u _ {x} = \sqrt {u ^ {2} - u _ {y} ^ {2}}} \\ & {\qquad = \sqrt {(1 2 5) ^ {2} - (1 0 0) ^ {2}}} \\ & {\qquad = 7 5 \mathrm{ms} ^ {- 1}} \end{array}$
Time taken to reach the top of the hill,
$t = \sqrt {\frac {2 h}{g}} = \sqrt {\frac {2 \times 5 0 0}{1 0}} = 1 0 \mathrm{s}$
Time taken to reach the ground from the top of the hill,
$t ^ {\prime} = t = 1 0 \mathrm{s}$
Horizontal distance travelled in 10 s,
$\begin{array}{r} x = u _ {x} \times t = 7 5 \times 1 0 \\ = 7 5 0 \mathrm{m} \end{array}$
∴ Distance through which canon has to be moved
$= 8 0 0 - 7 5 0 = 5 0 \mathrm{m}$
Speed with which canon can move $= 2\mathrm{ms}^{-1}$
∴ Time taken by canon, $t'' = \frac{50}{2}$ $\Rightarrow$ $= 25 \, s$
∴ Total time taken by a packet to reach on the ground $= t'' + t + t'$ $= 25 + 10 + 10 = 45 \, s$

$(d)\left|\frac{dv}{dt}\right|=|a|=9.8\ \mathrm{ms}^{-2}=$ constant and it is true that in case of projectile motion, the magnitude of velocity first decreases and then increases during the motion.

At highest point, velocity is horizontal and acceleration is vertical, i.e. both are perpendicular to each other and hence, their dot product is zero.

$h_{1}=\frac{u^{2}}{2g}, h_{2}=\frac{u^{2}\sin^{2}30^{\circ}}{2g}=\frac{u^{2}}{8g}$
$\Rightarrow h _ {1} = 4 h _ {2}$
Also, at highest point, v = 0 in first case
and $v \neq 0$ in second case.

For both cases, $t = \sqrt{\frac{2h}{g}} = constant.$
Because, initial vertical downward component of velocity will be zero for both the particles and both move under the effect of g.

Vertical component of velocity of ball at point P.
$\begin{array}{r} u _ {V} = 0 + g t \\ = 1 0 \times 0. 4 = 4 \mathrm{ms} ^ {- 1} \end{array}$
Horizontal component of velocity = initial velocity
$\Rightarrow v_{H} = 4\mathrm{m / s}$

So, the speed with which it hits the ground,
$v = \sqrt {v _ {H} ^ {2} + v _ {V} ^ {2}} = 4 \sqrt {2} \mathrm{m/s}$
and $\tan\theta=\frac{v_{V}}{v_{H}}=\frac{4}{4}=1\Rightarrow\theta=45^{\circ}$
It means the ball hits the ground at an angle of $45^{\circ}$ to the horizontal.
Height of the table $h=\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times(0.4)^{2}=0.8\ m$
Horizontal distance travelled by the ball from the edge of table $h = ut = 4 \times 0.4 = 1.6 \, m$

The angle between instantaneous velocity vector and acceleration vector before attaining the maximum height is acute (0 to $\pi/2$ ) and after is obtuse ( $\pi/2$ to $\pi$ ). At the highest point, it is perpendicular ( $\pi/2$ ).

$\begin{array}{l} t _ {1} = t _ {2} = \sqrt {\frac {2 h}{g}} \\ v _ {1} = \sqrt {2 g h} \end{array} \qquad \text {(where, h = height of tower)}$
While $v_{2}=\sqrt{v_{1}^{2}+v_{0}^{2}}$ (where, $v_{0}=initial horizontal velocity$ )
Therefore, both statements are correct.