Kinematics-2D

The motion of the object shot in two cases can be depicted as below


Case II
Using third equation of motion, $v^{2}=u^{2}-2gh$
As the object stops finally, so v = 0
For inclined motion, $g = g \sin \theta$ and h = x
Substituting these values in Eq. (i), we get
$u ^ {2} = 2 g \sin \theta x$
⇒
$x = \frac {u ^ {2}}{2 g \sin \theta}$
For case I $x_{1}=\frac{u^{2}}{2g\sin60^{\circ}}$
For case II $x_{2} = \frac{u^{2}}{2g\sin30^{\circ}}$
:
$\begin{array}{c} \frac {x _ {1}}{x _ {2}} = \frac {u ^ {2}}{2 g \sin 6 0 ^ {\circ}} \times \frac {2 g \sin 3 0 ^ {\circ}}{u ^ {2}} \\ = \frac {1}{2} \times \frac {2}{\sqrt {3}} = \frac {1}{\sqrt {3}} \text {or} 1: \sqrt {3} \end{array}$

$0 = \frac {u ^ {2}}{1 0 g} \mathrm{or} \frac {u ^ {2}}{g} = 2 0 0$
The range, $R = \frac{u^2 \times 2\sin\theta \cos\theta}{g} = \frac{200 \times 2 \times 2}{\sqrt{5} \times \sqrt{5}}$ $= \frac{200 \times 4}{5} = 160 \, \text{m}$

To obtain maximum range, angle of projection must be $45^{\circ}$ , i.e. $\theta = 45^{\circ}$ .
$R _ {\mathrm{max}} = \frac {u ^ {2} \sin (2 \times 4 5 ^ {\circ})}{g} = \frac {u ^ {2}}{g}$
:
$H _ {\mathrm{max}} = \frac {u ^ {2} \sin^ {2} 4 5 ^ {\circ}}{2 g} = \frac {u ^ {2}}{4 g} = \frac {R _ {\mathrm{max}}}{4}$
So, $H_{\mathrm{max}}$ is $25\%$ of $R_{\mathrm{max}}$ .

Given, $u = 98 \, ms^{-1}$ and $\theta = 30^{\circ}$
∵ Range of a projectile,
$\begin{array}{l} {R = \frac {u ^ {2} \sin (2 \theta)}{g} = \frac {9 8 \times 9 8 \times \sin 6 0 ^ {\circ}}{9 . 8}} \\ {R = 4 9 0 \sqrt {3} \mathrm{m}} \end{array}$

Horizontal range, $R = \frac{u^{2} \sin 2\theta}{g}$
When $\theta = 45^{\circ}$ , maximum horizontal range,
$R _ {\mathrm{max}} = \frac {u ^ {2}}{g} \sin 9 0 ^ {\circ} = \frac {u ^ {2}}{g}$
When $\theta = 135^{\circ}$ , maximum horizontal range,
$R = \frac {u ^ {2}}{g} \sin 2 7 0 ^ {\circ} = \frac {- u ^ {2}}{g}$
Negative sign implies opposite direction.

Given, $x = a \cos(pt)$ , $y = b \sin(pt)$
$\therefore \frac{x^2}{a^2} +\frac{y^2}{b^2} = 1$ , i.e. equation of ellipse
Now, $\mathbf{r} = x\hat{\mathrm{i}} +y\hat{\mathrm{j}} = a\cos (pt)\hat{\mathrm{i}} +b\sin (pt)\hat{\mathrm{j}}$
$\mathbf {v} = \frac {d \mathbf {r}}{d t} = - p a \sin (p t) \hat {\mathbf {i}} + p b \cos (p t) \hat {\mathbf {j}}$
$\mathbf {a} = \frac {d \mathbf {v}}{d t} = - p ^ {2} a \cos (p t) \hat {\mathbf {i}} - p ^ {2} b \sin (p t) \hat {\mathbf {j}}$
At
$\begin{array}{l} {t = \frac {\pi}{2 p},} \\ {{\bf v} = - p a \hat {\bf i} \mathrm{and} {\bf a} = - p ^ {2} b \hat {\bf j}} \end{array}$
Thus, velocity is perpendicular to acceleration.
Also, $a = -p^{2}b$ , i.e. directed towards a fixed point as p and b are positive constants.
$\mathrm{As}, \qquad v = \frac {d r}{d t} \Rightarrow \Delta r = \int_ {0} ^ {\frac {\pi}{2 p}} v d t$
$\Rightarrow \Delta r = [ a \cos (p t) \hat {\mathbf {i}} + b \sin (p t) \hat {\mathbf {j}} ] _ {0} ^ {\frac {\pi}{2 p}} = - a \hat {\mathbf {i}} + b \hat {\mathbf {j}}$
$\mathrm{So}, \qquad | \mathbf {r} | = \sqrt {a ^ {2} + b ^ {2}}$

Position vector of the particle is given by $\mathbf{r} = \cos \omega t\hat{\mathbf{x}} +\sin \omega t\hat{\mathbf{y}}$
where, $\omega$ is a constant.
Velocity of the particle is
$\begin{array}{r l} & {\mathbf {v} = \frac {d \mathbf {r}}{d t} = \frac {d}{d t} (\cos \omega t \hat {\mathbf {x}} + \sin \omega t \hat {\mathbf {y}})} \\ & {\quad = (- \sin \omega t) \omega \hat {\mathbf {x}} + (\cos \omega t) \omega \hat {\mathbf {y}}} \\ & {\quad = - \omega (\sin \omega t \hat {\mathbf {x}} - \cos \omega t \hat {\mathbf {y}})} \end{array}$
Acceleration of the particle,
$\begin{array}{r l} & {\mathbf {a} = \frac {d \mathbf {v}}{d t} = \frac {d}{d t} (- \boldsymbol {\omega} \sin \boldsymbol {\omega} t \hat {\mathbf {x}} + \boldsymbol {\omega} \cos \boldsymbol {\omega} t \hat {\mathbf {y}})} \\ & {\quad = - \boldsymbol {\omega} ^ {2} \cos \boldsymbol {\omega} t \hat {\mathbf {x}} - \boldsymbol {\omega} ^ {2} \sin \boldsymbol {\omega} t \hat {\mathbf {y}}} \end{array}$
$\Rightarrow \mathbf {a} = - \omega^ {2} \mathbf {r} = \omega^ {2} (- \mathbf {r})$
Assuming the particle is at P, then its position vector is directed as shown in the diagram.

Therefore, acceleration is directed towards -r, i.e. towards O (origin).
$\mathbf{v}\cdot \mathbf{r} = -\boldsymbol {\omega}(\sin \omega t\hat{\mathbf{x}} -\cos \omega t\hat{\mathbf{y}})\cdot (\cos \omega t\hat{\mathbf{x}} +\sin \omega t\hat{\mathbf{y}})\\ = -\boldsymbol {\omega}\left[\sin \omega t\cdot \cos \omega t + 0 + 0 - \sin \omega t\cdot \cos \omega t\right]\\ = -\boldsymbol {\omega}(0) = 0$
$\Rightarrow \mathbf{v} \perp \mathbf{r}$
Thus, velocity is perpendicular to r.

The equation of trajectory,
$y = x \tan \alpha \left[ 1 - \frac {x}{R} \right]\tag{Gives}$
$Q = P \tan \theta \left[ 1 - \frac {P}{R} \right]\tag{... (i}$
and
$P = Q \tan \theta \left[ 1 - \frac {Q}{R} \right]\tag{... (ii}$
On dividing Eq. (i) by Eq. (ii), we get
$\frac {Q ^ {2}}{P ^ {2}} = \frac {[ 1 - P / R ]}{[ 1 - Q / R ]}$
$\frac {1}{R} [ P ^ {3} - Q ^ {3} ] = P ^ {2} - Q ^ {2}$
⇒
$R = \frac {P ^ {3} - Q ^ {3}}{P ^ {2} - Q ^ {2}} = \frac {P ^ {2} + P Q + Q ^ {2}}{P + Q}$
Now,
$\begin{array}{r} \frac {Q}{P} = \tan \theta \left[ 1 - \frac {P (P + Q)}{P ^ {2} + P Q + Q ^ {2}} \right] \\ = \tan \theta \left[ \frac {P ^ {2} + P Q + Q ^ {2} - P ^ {2} - P Q}{P ^ {2} + P Q + Q ^ {2}} \right] \end{array}$
⇒
$\tan \theta = \frac {P ^ {2} + Q ^ {2} + P Q}{P Q}$
⇒
$\theta = \tan^ {- 1} \left[ \frac {P ^ {2} + P Q + Q ^ {2}}{P Q} \right]$

Consider the diagram

For horizontal motion, $50 = 10\cos 30^{\circ}\times t$
⇒
$t = \frac {5}{\cos 3 0 ^ {\circ}} = \frac {5}{\frac {\sqrt {3}}{2}} = \frac {1 0}{\sqrt {3}}$
For vertical motion,
$\begin{array}{r l} & h = 1 0 \sin 3 0 ^ {\circ} \times \frac {1 0}{\sqrt {3}} - \frac {1}{2} \times 1 0 \times \left(\frac {1 0}{\sqrt {3}}\right) ^ {2} \\ & \quad = 1 0 \times \frac {1}{2} \times \frac {1 0}{\sqrt {3}} - \frac {1 0 0}{\sqrt {3}} \times \frac {5}{\sqrt {3}} = \frac {1 0 0}{\sqrt {3}} \left(\frac {1}{2} - \frac {5}{\sqrt {3}}\right) \\ & \quad = \frac {5 0}{\sqrt {3}} \left[ 1 - \frac {1 0}{\sqrt {3}} \right] \mathrm{m} \end{array}$

Range of projectile is same for $\theta$ and $90^{\circ}-\theta$ .
Given, $\theta = 40^{\circ}$
Another angle $= 90^{\circ} - \theta = 90^{\circ} - 40^{\circ} = 50^{\circ}$

Given, a particle having horizontal range is twice the greatest height attained by it, i.e. R = 2H
$\frac {v ^ {2} \sin 2 \theta}{g} = \frac {2 v ^ {2} \sin^ {2} \theta}{2 g}$

$\Rightarrow 2\sin \theta \cos \theta = \sin^{2}\theta \Rightarrow \tan \theta = 2$
$\Rightarrow \sin \theta = \frac {2}{\sqrt {5}} \Rightarrow \cos \theta = \frac {1}{\sqrt {5}}$
$\therefore R = \frac{v^2\sin 2\theta}{g} = \frac{v^2\times 2\sin\theta\cos\theta}{g}$
$\Rightarrow R = v^{2}\times \frac{2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$
i.e. $R = \frac{4v^2}{5g}$

Given, $\theta_{1} = \theta_{2} = 45^{\circ}$
$\theta_ {1} + \theta_ {2} = (4 5 ^ {\circ} + \alpha) + (4 5 ^ {\circ} - \alpha) = 9 0 ^ {\circ}$
$u _ {1} = u _ {2} = u \Rightarrow R _ {1} = R _ {2} = ?$
$R _ {1} = \frac {u _ {1} ^ {2} \sin (\theta_ {1} + \theta_ {2})}{g}$
⇒
$R _ {1} = \frac {u ^ {2} \sin 9 0 ^ {\circ}}{g} = \frac {u ^ {2}}{g}\tag{... (i}$
Similarly, $R_{2} = \frac{u_{2}^{2}\sin(\theta_{1} + \theta_{2})}{g}$
⇒
$R _ {2} = \frac {u ^ {2} \sin 9 0 ^ {\circ}}{g} = \frac {u ^ {2}}{g}\tag{... (ii}$
The ratio of horizontal ranges
$\frac {R _ {1}}{R _ {2}} = \frac {u ^ {2} / g}{u ^ {2} / g} = 1: 1 \Rightarrow R _ {1}: R _ {2} = 1: 1$

Average velocity, $v_{av} = \frac{Net displacement}{Time taken}$ $= \frac{(13 - 2)\hat{i} + (14 - 3)\hat{j}}{5}$ $= \frac{11\hat{i} + 11\hat{j}}{5} = \frac{11}{5} (\hat{i} + \hat{j})$

For vertical motion, $h_{1}=u\sin\theta t_{1}-\frac{1}{2}gt_{1}^{2}$
(For $h_{1}$ )
$t _ {1} = \frac {h _ {1} + \frac {1}{2} g t _ {1} ^ {2}}{u \sin \theta}\tag{... (i}$
⇒
$h _ {2} = u \sin \theta t _ {2} - \frac {1}{2} g t _ {2} ^ {2}\tag{For \(h_2\)}$
$t _ {2} = \frac {h _ {2} + \frac {1}{2} g t _ {2} ^ {2}}{u \sin \theta}$
⇒
... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac {t _ {1}}{t _ {2}} = \frac {\left(h _ {1} + \frac {1}{2} g t _ {1} ^ {2}\right) / u \sin \theta}{\left(h _ {2} + \frac {1}{2} g t _ {2} ^ {2}\right) / u \sin \theta}$
$\Rightarrow h _ {1} t _ {2} - h _ {2} t _ {1} = \frac {g}{2} (t _ {1} t _ {2} ^ {2} - t _ {1} ^ {2} t _ {2})$
The time of flight of the ball,
$\begin{array}{r l r} & {T = \frac {2 u \sin \theta}{g} = \frac {2}{g} (u \sin \theta)} \\ & {\quad = \frac {2}{g} \left(\frac {h _ {1} + 1 / 2 g t _ {1} ^ {2}}{t _ {1}}\right)} & {[ \mathrm{FromEq.(i)} ]} \\ & {\quad = \frac {2}{t _ {1}} \left(\frac {h _ {1}}{g} + \frac {t _ {1} ^ {2}}{2}\right)} \\ & {\quad = \frac {h _ {1}}{t _ {1}} \times \frac {2}{g} + t _ {1} = \frac {h _ {1}}{t _ {1}} \times \left(\frac {t _ {1} t _ {2} ^ {2} - t _ {1} ^ {2} t _ {2}}{h _ {1} t _ {2} - h _ {2} t _ {1}}\right) + t _ {1}} \\ & {\quad = \frac {h _ {1} t _ {1} t _ {2} ^ {2} - h _ {1} t _ {1} ^ {2} t _ {2} + h _ {1} t _ {1} ^ {2} t _ {2} - h _ {2} t _ {1} ^ {3}}{t _ {1} (h _ {1} t _ {2} - h _ {2} t _ {1})} = \left(\frac {h _ {1} t _ {2} ^ {2} - h _ {2} t _ {1} ^ {2}}{h _ {1} t _ {2} - h _ {2} t _ {1}}\right)} \end{array}$

As we know that, maximum height of a projectile is given by
$H _ {\mathrm{max}} = \frac {u ^ {2} \sin^ {2} \theta}{2 g}$
where, u = initial velocity of projectile
g = acceleration due to gravity
and $\theta = \text{angle of projection.}$
As per question,
$H _ {\mathrm{max}} = \frac {v ^ {2} \sin^ {2} 4 5 ^ {\circ}}{2 g}$
$\dots (\mathrm{i}) (\mathrm{As}, u = v, \theta = 4 5 ^ {\circ})$
Now, range of a projectile is given by
$R = \frac {u ^ {2} \sin 2 \theta}{g}$
⇒
$R = \frac {v ^ {2} \sin (2 \times 4 5 ^ {\circ})}{g}$
⇒
$R = \frac {v ^ {2} \sin 9 0 ^ {\circ}}{g}\tag{... (ii}$
On dividing Eq. (i) by Eq. (ii), we get
$\frac {H _ {\mathrm{max}}}{R} = \frac {v ^ {2} \sin^ {2} 4 5 ^ {\circ} \times g}{2 g \times v ^ {2} \sin 9 0 ^ {\circ}} = \frac {1}{4 \times 1}$

The horizontal component of velocity,
$v _ {x} = u = 1 0 \mathrm{ms} ^ {- 1}$

Now, $\tan \theta = \tan 45^{\circ} = \frac{v_y}{v_x} = \frac{v_y}{10}$
$\Rightarrow \quad v _ {y} = 1 0 \mathrm{ms} ^ {- 1}\tag{18 (b}$
$b) \mathrm{As}, \qquad t = \frac {2 u \sin \theta}{g} \Rightarrow 4 = \frac {2 u \sin \theta}{g}\tag{... (i}$

$\Rightarrow u\sin \theta = 2g$
and $h=\frac{u^{2}\sin^{2}\theta}{2g}=\frac{4g^{2}}{2g}$
[From Eq.(i)]
$\Rightarrow h = 20 \mathrm{~m}$

B

From the question's figure, the $x$ -component remain unchanged, while the $y$ -component is reverse. Then, the velocity at point $B$ is $(2\hat{\mathbf{i}} - 3\hat{\mathbf{j}})\mathrm{ms}^{-1}$ .

Time of flight, $T = \frac{2u \sin \theta}{g}$
Given, $u = u_{0}$ , $T = 1\mathrm{s}$ , $\theta = 30^{\circ}$ , $g = 9.8\mathrm{ms}^{-2}$
$u _ {0} = g = 9. 8 \mathrm{ms} ^ {- 1}$

As the velocity makes an angle of $30^{\circ}$ with horizontal, so the horizontal component of velocity at the instant will be $v \cos 30^{\circ}$ .
$\Rightarrow v\cos 30^{\circ} = 5$
$\Rightarrow v = \frac {5}{\cos 3 0 ^ {\circ}} = \frac {5}{\sqrt {3} / 2} = \frac {1 0}{\sqrt {3}} \mathrm{ms} ^ {- 1}$

As the horizontal ranges are the same.
$\therefore \frac {v _ {0} ^ {2} \sin 2 \theta_ {1}}{g} = \frac {v _ {0} ^ {2} \sin 2 \theta_ {2}}{g}$
$\mathrm{So}, \quad \sin 2 \theta_ {1} = \sin 2 \theta_ {2}$
$\text { or } \qquad 2 \theta_ {1} = \pi - 2 \theta_ {2}$
$\Rightarrow \theta_ {1} + \theta_ {2} = \pi / 2$
$(h _ {1}) _ {\mathrm{max}} = \frac {v _ {0} ^ {2} \sin^ {2} \theta_ {1}}{2 g}$
and
$(h _ {2}) _ {\mathrm{max}} = \frac {v _ {0} ^ {2} \sin^ {2} \theta_ {2}}{2 g}$
:
$\begin{array}{r l} \frac {(h _ {1}) _ {\max}}{(h _ {2}) _ {\max}} = & \frac {\sin^ {2} \theta_ {1}}{\sin^ {2} \theta_ {2}} = \frac {\sin^ {2} \theta_ {1}}{\cos^ {2} \theta_ {1}} = \tan^ {2} \theta_ {1} \\ & \left[ \because \sin^ {2} \theta_ {2} = \sin^ {2} \left(\frac {\pi}{2} - \theta_ {2}\right) = \cos^ {2} \theta / 2 \right] \end{array}$

Given, $r = 2t \hat{i} + t^{2} \hat{j}$
Velocity vector, $\mathbf{v} = \frac{d\mathbf{r}}{dt}$ and using $\frac{d}{dx} x^n = nx^{n - 1}$
We have, $\frac{d\mathbf{r}}{dt} = 2\hat{\mathbf{i}} + 2t\hat{\mathbf{j}}$

Given, R = H
$\mathrm{Range}, R = \frac {u ^ {2} (2 \sin \theta \cos \theta)}{g}$
$\mathrm{Height}, H = \frac {u ^ {2} \sin^ {2} \theta}{2 g}$
$\mathrm{Hence}, \frac {u ^ {2} (2 \sin \theta \cos \theta)}{g} = \frac {u ^ {2} \sin^ {2} \theta}{2 g}$

Maximum height and time of flight depends upon the vertical component of initial velocity.
$\begin{array}{l l} & H _ {1} > H _ {2} \\ \Rightarrow & u _ {y _ {1}} > u _ {y _ {2}} \\ \text { Range } & R _ {2} > R _ {1} \\ \text { So, } & u _ {2} > u _ {1} \end{array}$

According to given condition,
$\frac {h _ {1}}{h _ {2}} = \frac {\sin^ {2} \theta}{\sin^ {2} (9 0 ^ {\circ} - \theta)} = \frac {2 0}{8 0}$
$\Rightarrow \quad \tan^ {2} \theta = \frac {1}{4} \Rightarrow \tan \theta = \frac {1}{2}$
$\sin \theta = \frac {1}{\sqrt {5}} \mathrm{and} \cos \theta = \frac {2}{\sqrt {5}}$
So,
$h = \frac {u ^ {2} \sin^ {2} \theta}{2 g}$
⇒