(a) the effective angle to the horizontal at which the ball is projected inair as seen by a spectator.
(b) what will be time of flight?
(c) what is the distance (horizontal range) from the point of projection at which the ball will land?
(d) find $\theta$ at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) how does $\theta$ for maximum range change if $u > u_0, u = u_0, u < v_0$?
(f) how does $\theta$ in (v) compare with that for $u = 0 \ (i.e., 45^\circ)$?
Consider the adjacent diagram.
(a) Initial velocity in
x-direction,
$u_x = u + v_0 \cos\theta$
$u_y = \text{Initial velocity in y-direction}$
$= v_0 \sin\theta$
where angle of projection is $\theta$.
Now, we can write
$\tan\theta = \frac{u_y}{u_x} = \frac{v_0 \sin\theta}{u + v_0 \cos\theta}$
$\Rightarrow \theta = \tan^{-1} \left( \frac{v_0 \sin\theta}{u + v_0 \cos\theta} \right)$
(b) Let T be the time of flight.
As net displacement is zero over time period T.
$ y = 0,\ u_y = v_0 \sin \theta,\ a_y = -g,\ t = T $
We know that
$ y = u_y t + \frac{1}{2}a_y t^2 $
$ \Rightarrow \ 0 = v_0 \sin \theta \ T + \frac{1}{2}\ (-g)\ T^2 $
$ \Rightarrow \ T \left[ v_0 \sin \theta - \frac{g}{2} T \right] = 0 \ \Rightarrow \ \ T = 0, \frac{2v_0 \sin \theta}{g} $
$ T=0_1, \ \text{corresponds to point O.} $
Hence,$ T = \frac{2u_0 \sin \theta}{g} $
(c) Horizontal range, $ R = (u + v_0 \cos \theta) T = (u + v_0 \cos \theta) \frac{2v_0 \sin \theta}{g} $
$ = \frac{v_0}{g} \left[ 2u \sin \theta + v_0 \sin 2\theta \right] $
(d) For horizontal range to be maximum, $ \frac{dR}{d\theta} = 0 $
$ \Rightarrow \quad \frac{v_0}{g} \left[ 2u \cos \theta + v_0 \cos 2\theta \times 2 \right] = 0 $
$ \Rightarrow \quad 2u \cos \theta + 2v_0 \left[ 2\cos^2 \theta - 1 \right] = 0 $
$ \Rightarrow \quad 4v_0 \cos^2 \theta + 2u \cos \theta - 2v_0 = 0 $
$ \Rightarrow \quad 2v_0 \cos^2 \theta + u \cos \theta - v_0 = 0 $
$ \Rightarrow \quad \cos \theta = \frac{-u \pm \sqrt{u^2 + 8v_0^2}}{4v_0} $
$ \Rightarrow \quad \theta_{\text{max}} = \cos^{-1} \left[ \frac{-u \pm \sqrt{u^2 + 8v_0^2}}{4v_0} \right] $
(e) If $ u = v_0 $,
$ \cos \theta = \frac{-v_0 \pm \sqrt{v_0^2 + 8v_0^2}}{4v_0} = \frac{-1 + 3}{4} = \frac{1}{2} $
$ \theta = 60^\circ $
$ \Rightarrow $ If $ u \ll v_0 $, then $ 8v_0^2 + u^2 \approx 8v_0^2 $
$ \theta_{\max} = \cos^{-1} \left[ \frac{-u + 2 \sqrt{2}v_0}{4v_0} \right] = \cos^{-1} \left[ \frac{1}{\sqrt{2}} - \frac{u}{4v_0} \right] $
If $ u \ll v_0 $, then
$ \theta_{\max} = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = \frac{\pi}{4} $
If $ u > u_0 $ and $ u \gg v_0 $
$ \theta_{\max} = \cos^{-1} \left[ \frac{-u + u}{4v_0} \right] = 0 \Rightarrow \theta_{\max} = \frac{\pi}{2} $
(f) If $ u = 0 $, $ \theta_{\max} = \cos^{-1} \left[ \frac{0 \pm \sqrt{8v_0^2}}{4v_0} \right] = \cos^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45^\circ $
Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in cartesian coordinates $ \mathbf{A} = A_x \hat{i} + A_y \hat{j} $, where $ \hat{i} $ and $ \hat{j} $ are unit vector along x and y-directions, respectively and $ A_x $ and $ A_y $ are corresponding components of $ \mathbf{A} $. Motion can also be studied by expressing vectors in circular polar coordinates as $ \mathbf{A} = A_r \hat{r} + A_\theta \hat{\theta} $, where $ \hat{r} = \frac{\mathbf{r}}{r} = \cos{\theta} \hat{i} + \sin{\theta} \hat{j} $ and $ \hat{\theta} = -\sin{\theta} \hat{i} + \cos{\theta} \hat{j} $ are unit vectors along direction in which r and $ \theta $ are increasing.
(a) Express $ \hat{i} $ and $ \hat{j} $ in terms of $ \hat{r} $ and $ \hat{\theta} $.
(b) Show that both $ \hat{r} $ and $ \hat{\theta} $ are unit vectors and are perpendicular to each other.
(c) Show that $ \frac{d}{dt} ( \hat{r}) = \omega \hat{\theta} $, where $ \omega = \frac{d\theta}{dt} $ and $ \frac{d}{dt} ( \hat{\theta}) = - \omega \hat{r} $.
(d) For a particle moving along a spiral given by $ r = a \theta \hat{r} $, where $ a = 1 $ (unit), find dimensions of a.
(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.
(a) Given, unit vector
$ \hat{r} = \cos \theta \hat{i} + \sin \theta \hat{j} $ $ \hspace{2cm} \text{...(i)}$
$ \hat{\theta} = -\sin \theta \hat{i} + \cos \theta \hat{j} $$ \hspace{1.5cm} \text{...(ii)}$
Multiplying Eq. (i) by $ \sin \theta $ and Eq. (ii) with $ \cos \theta $ and adding
$ \hat{r} \sin \theta + \hat{\theta} \cos \theta = \sin \theta \cos \theta \hat{i} + \sin^2 \theta \hat{j} + \cos^2 \theta \hat{j} - \sin \theta \cos \theta \hat{i} $
$ = \hat{j} (\cos^2 \theta + \sin^2 \theta) = \hat{j} $
$ \Rightarrow \hat{r} \sin \theta + \hat{\theta} \cos \theta = \hat{j} $
By Eq. (i) $ \times \cos \theta - $ Eq. (ii) $ \times \sin \theta $
$ n (\hat{r} \cos \theta - \hat{\theta} \sin \theta) = \hat{i} $
(b) $ \mathbf{\hat{r}} \cdot \mathbf{\hat{\theta}} = (\cos \theta\; \mathbf{\hat{i}} + \sin \theta\; \mathbf{\hat{j}}) \cdot (- \sin \theta\; \mathbf{\hat{i}} + \cos \theta\; \mathbf{\hat{j}}) = - \cos \theta \cdot \sin \theta + \sin \theta \cdot \cos \theta = 0 $
$ \Rightarrow \theta = 90^\circ \; \text{Angle between} \; \mathbf{\hat{r}} \; \text{and} \; \mathbf{\hat{\theta}}. $
(c) Given, $ \mathbf{\hat{r}} = \cos \theta\; \mathbf{\hat{i}} + \sin \theta\; \mathbf{\hat{j}} $
$ \frac{d\mathbf{\hat{r}}}{dt} = \frac{d}{dt}(\cos \theta\; \mathbf{\hat{i}} + \sin \theta\; \mathbf{\hat{j}}) = - \sin \theta \cdot \frac{d\theta}{dt}\; \mathbf{\hat{i}} + \cos \theta \cdot \frac{d\theta}{dt}\; \mathbf{\hat{j}} $
$ = \omega [-\sin \theta\; \mathbf{\hat{i}} + \cos \theta\; \mathbf{\hat{j}}] $
$ \left(\therefore \; \theta = \frac{d\theta}{dt} \right) $
(d) Given, $ \mathbf{r} = a\theta\mathbf{\hat{r}}, $ here, writing dimensions $[\mathbf{r}] = [a][\theta][\mathbf{\hat{r}}] $
$ \Rightarrow L = [a] \% 1 \Rightarrow [a] = L = [M^0L^1T^0] $
(e) Given, $ a = 1 $ unit
$ r = \theta \hat{\mathbf{r}} = \theta [\cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}}] $
Velocity,
$ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d\theta}{dt} \hat{\mathbf{r}} + \theta \frac{d}{dt} \hat{\mathbf{r}} = \frac{d\theta}{dt} \hat{\mathbf{r}} + \theta \frac{d}{dt} [(\cos \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}})] $
$ = \frac{d\theta}{dt} \hat{\mathbf{r}} + \theta \left[ (-\sin \theta \hat{\mathbf{i}} + \cos \theta \hat{\mathbf{j}}) \frac{d\theta}{dt} \right] $
$ = \frac{d\theta}{dt} \hat{\mathbf{r}} + \theta \hat{\mathbf{\theta}} = \omega \hat{\mathbf{r}} + \omega \hat{\mathbf{\theta}} $
Acceleration,
$ \mathbf{a} = \frac{d}{dt} [\omega \hat{\mathbf{r}} + \omega \theta\hat{\mathbf{\theta}}] = \frac{d}{dt} \left[ \frac{d\theta}{dt} \hat{\mathbf{r}} + \frac{d\theta}{dt} (\theta \hat{\mathbf{\theta}}) \right] $
$ = \frac{d^2\theta}{dt^2} \hat{\mathbf{r}} + \frac{d\theta}{dt} \cdot \frac{d\hat{\mathbf{r}}}{dt} + \frac{d^2\theta}{dt^2} \hat{\mathbf{\theta}} + \omega \frac{d}{dt} (\theta \hat{\mathbf{\theta}}) $
$ = \frac{d^2\theta}{dt^2} \hat{\mathbf{r}} + \omega [-\sin \theta \hat{\mathbf{i}} + \sin \theta \hat{\mathbf{j}}] + \frac{d^2\theta}{dt^2} \hat{\mathbf{\theta}} + \omega \frac{d}{dt} (\theta \hat{\mathbf{\theta}}) $
$ = \frac{d^2\theta}{dt^2} \hat{\mathbf{r}} + \omega^2 \hat{\mathbf{\theta}} + \frac{d^2\theta}{dt^2} \times \hat{\mathbf{\theta}} + \omega^2 \hat{\mathbf{\theta}} + \omega^2 (\theta \hat{\mathbf{r}}) $
$ = \left( \frac{d^2\theta}{dt^2} - \omega^2 \right) \hat{\mathbf{r}} + \left(2\omega^2 + \frac{d^2\theta}{dt^2} \theta \right) \hat{\mathbf{\theta}} $
A man wants to reach from A to the opposite corner of the square C. The sides of the square are 100 m. A central square of 50m × 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed v m/s (v < 1). What is the smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?
Consider adjacent diagram.
Time taken to go from A to C via straight line path APQC through the S and
$ T_{\text{sand}} = \frac{AP + QC}{1} + \frac{PQ}{v} = \frac{25\sqrt{2} + 25\sqrt{2}}{1} + \frac{50\sqrt{2}}{v}\\ = 50\sqrt{2} + \frac{50\sqrt{2}}{v} = 50\sqrt{2} \left( \frac{1}{v} + 1 \right) $
Clearly from figure the shortest path outside the sand will be ARC.
Time taken to go from A to C via this path
$ T_{\text{outside}} = \frac{AR + RC}{1} $
Clearly,
$ AR = \sqrt{75^2 + 25^2} = \sqrt{75 \times 75 + 25 \times 25}\\ = 5 \times 5\sqrt{9 + 1} = 25\sqrt{10} \text{ m} $
$ RC = AR = \sqrt{75^2 + 25^2} = 25\sqrt{10} \text{ m} $
$\Rightarrow T_{\text{outside}} = 2AR = 2 \times 25\sqrt{10} \text{ s} = 50\sqrt{10} \text{ s} $
For
$ T_{\text{sand}} < T_{\text{outside}} $
$\Rightarrow 50\sqrt{2}\left( \frac{1}{v} + 1 \right) < 2 \times 25\sqrt{10} $
$\Rightarrow \frac{2\sqrt{2}}{2}\left( \frac{1}{v} + 1 \right) < \sqrt{10} $
$\Rightarrow \frac{1}{v} + 1 < \frac{2\sqrt{10}}{2\sqrt{2}} = \frac{\sqrt{5}}{2} \times 2 = \sqrt{5} $
$\Rightarrow \frac{1}{v} < \frac{\sqrt{5}}{2} \times 2 - 1 \ \ \Rightarrow \ \ \frac{1}{v} < \sqrt{5} - 1 $
$\Rightarrow v > \frac{1}{\sqrt{5} - 1} \approx 0.81 \text{m/s} $
$\Rightarrow v > 0.81 \text{m/s} $