Kinematics-2D

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.

Let at any instant of time, the length AB be l, here angle $\theta$ and length l vary with time, then using Pythagoras theorem in $\Delta ABC$ ,

$x ^ {2} + y ^ {2} = l ^ {2}$
On differentiating both sides w.r.t t, we get
or

$\left|\Delta\vec{v}\right|=\sqrt{v^{2}+v^{2}-2v^{2}\cos\theta}$ $=2v\sin\theta/2$ .

The minimum possible velocity at B is, $V_{B} = \sqrt{g\ell}$
Then the range for this is,
$R = V _ {B} \sqrt {\frac {2 h}{g}} = \sqrt {g \ell} \times \sqrt {\frac {2 (2 \ell)}{g}} = 2 \ell .$

$y = x \tan \theta - \frac{gx^2}{2u^2 \cos^2 \theta}$ For equal trajectories for same angle of projection $\frac{g}{u^2} = \text{constant}$ $\Rightarrow \frac{9.8}{5^2} = \frac{g'}{3^2}$ $g' = \frac{9.8 \times 9}{25} = 3.528m/s^2 = 3.5m/s^2.$

$f _ {L} = \mu N = \mu m r \omega^ {2}$
$f _ {s} = m g$
$\text { As } f _ {s} \leq f _ {L}$
$\Rightarrow m g \leq \mu m r \omega^ {2}$
$\Rightarrow \omega \geq \sqrt {\frac {g}{\mu r}}$

$\Rightarrow \omega_ {\min} = 1 0 r a d / s.$

$\frac{h_1}{d} = \tan 45^\circ \Rightarrow h_1 = d$ ... (i)
$\frac {h _ {1} + h _ {2}}{d + 2 . 4 6 4 d} = \tan 3 0 ^ {\circ}$
$\Rightarrow (h _ {1} + h _ {2}) \times \sqrt {3} = 3. 4 6 4 d$

$(h _ {1} + h _ {2}) = \frac {3 . 4 6 4 d}{\sqrt {3}}$
$\Rightarrow d + h _ {2} = \frac {3 . 4 6 4 d}{\sqrt {3}} \Rightarrow h _ {2} = d.$

At the highest point, vertical component of velocity becomes zero so there will be only horizontal velocity and it is perpendicular to the acceleration due to gravity.

$H = \frac{u^2\sin^2\theta}{2g}$ i.e. it is independent of mass of projectile.

$R = \frac{u^2\sin 2\theta}{g}\therefore R_{\max} = \frac{u^2}{g}$
when $\theta = 45^{\circ}\therefore R_{\max}\propto u^{2}$
Height $H = \frac{u^2\sin^2\theta}{2g}$
$\Rightarrow H _ {\max} = \frac {u ^ {2}}{2 g}$
when $\theta = 90^{\circ}$
It is clear that $H_{\max} = \frac{R_{\max}}{2}$ .

By doing so component of weight of vehicle provides centripetal force.