Motion in a Plane

The given situation is shown below :

According to given condition, range must be greater than 20 m and less than 24.2 m .

$ \Rightarrow \quad 20 \leq R \leq 24.2 $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 20 \leq \frac{u^2 \sin 2 \theta}{g} \leq 24.2 \\ & \Rightarrow \quad 20 \leq \frac{u^2 \times \sin \left(2 \times 15^{\circ}\right)}{10} \leq 24.2 \\ & \Rightarrow \quad 200 \leq u^2 \times 1 / 2 \leq 242\left(\because \sin 30^{\circ}=\frac{1}{2}\right) \\ & \Rightarrow \quad 400 \leq u^2 \leq 484 \\ & \Rightarrow \quad 20 \mathrm{~m} / \mathrm{s} \leq u \leq 22 \mathrm{~m} / \mathrm{s} \end{aligned}\\ &\text { So, option (a) is best choice. } \end{aligned} $

The range of two particles are same, that means angle of projections must be complementary to each other.

So one angle = $\theta $ and other one is = 90^o - $\theta $

R = ${{{u^2}\sin 2\theta } \over g}$ = ${{2{u^2}\sin \theta \cos \theta } \over g}$

$ \therefore $ R² = ${{4{u^4}{{\sin }^2}\theta {{\cos }^2}\theta } \over {{g^2}}}$

h₁ = ${{{u^2}{{\sin }^2}\theta } \over {2g}}$

h₂ = ${{{u^2}{{\sin }^2}\left( {{{90}^o} - \theta } \right)} \over {2g}}$ = ${{{u^2}{{\cos }^2}\theta } \over {2g}}$

h₁h₂ = ${{{u^4}{{\sin }^2}\theta {{\cos }^2}\theta } \over {4{g^2}}}$

$ \Rightarrow $ h₁h₂ = ${{{R^2}} \over {16}}$

$ \Rightarrow $ R² = 16 h₁h₂

Range will be same for time t₁ and t₂, so angles of projection will be ‘$\theta $’ & ‘90° – $\theta $’

${t_1} = {{2u\sin \theta } \over g}{t_2} = {{2u\sin \left( {{{90}^o} - \theta } \right)} \over g}$

and $R = {{{u^2}\sin 2\theta } \over g}$

${t_1}{t_2} = {{4{u^2}\sin \theta \cos \theta } \over {{g^2}}} = {2 \over g}\left[ {{{2{u^2}\sin \theta \cos \theta } \over g}} \right]$

= ${{2R} \over g}$

Equation of trajectory is given as
y = 2x – 9x²     …(A)

Comparing with equation:
$y = x\tan \theta - {g \over {2{u^2}{{\cos }^2}\theta }}{x^2}$    ...(B)

We get, $\tan \theta = 2$

$ \therefore \cos \theta = {1 \over {\sqrt 5 }}$

Also, ${g \over {2{u^2}{{\cos }^2}\theta }} = 9$

$ \Rightarrow {{10} \over {2 \times 9 \times {{\left( {{1 \over {\sqrt 5 }}} \right)}^2}}} = {u^2};\,\,{u^2} = {{25} \over 9}$

$ \Rightarrow u = {5 \over 3}m/s$

$T = {{2u\sin \theta } \over {g\cos \alpha }}$

$R = u\cos \theta T - {1 \over 2}g\sin \alpha {T^2}$

$ = {{u\cos \theta 2u\sin \theta } \over {g\cos \alpha }} - {{g\sin \alpha } \over 2}{{4{u^2}{{\sin }^2}\theta } \over {{g^2}{{\cos }^2}\alpha }}$

$ = {{{u^2}{{\sin }^2}\theta } \over {g\cos \alpha }} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - \cos 2\theta } \right\}$

$ = {{4 \times {1 \over 2}} \over {10 \times {{\sqrt 3 } \over 2}}} - {{{u^2}\sin \alpha } \over {g{{\cos }^2}\alpha }}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$

$ = {4 \over {10\sqrt 3 }} - {8 \over {30}}\left\{ {1 - {{\sqrt 3 } \over 2}} \right\}$

$ = {4 \over {5\sqrt 3 }} - {8 \over {30}} = {{8\sqrt 3 - 8} \over {30}}$

$ = {{8\left( {\sqrt 3 - 1} \right)} \over {30}} = 20\,cm$

Draw velocity diagram
$\sin \theta = {{{v_r}} \over {{v_{sr}}}} = {1 \over 2}$

$\theta = {30^o}$

$\phi $ = 90 + $\theta $ = 120°

Considering the initial position of ship A as origin, so the velocity and position of ship will be

${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$ and ${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$

Now, as given in the question, velocity and position of ship B will be, ${\overrightarrow v _B} = - 10\widehat i$ and ${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$

Time after which the distance is minimum between A and B can be calculated as

$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$

where, ${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$

and ${\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)$

$ = - 40\widehat i - 50\widehat j$

$ \Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$

$ = {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$ h

AB = V_P $ \times $ t

BC = Vt

cos60^o = ${{AB} \over {BC}}$

${1 \over 2} = {{{V_P} \times t} \over {Vt}}$

V_P = ${V \over 2}$

$R = {{{u^2}\sin 2\theta } \over g}$

$A = \pi \,{R^2}$

$A \propto {R^2}$

$A \propto {u^4}$

${{{A_1}} \over {{A_2}}} = {{u_1^4} \over {u_2^4}} = {\left[ {{1 \over 2}} \right]^4} = {1 \over {16}}$

Given that,

x = a cos $\omega $t

y = a sin $\omega $t

z = a $\omega $t

Velocity in x-direction,

V_x = ${{dx} \over {dt}} = - a\omega \sin \omega t$

Velocity in y-direction,

V_y = ${{dy} \over {dt}}$ = a $\omega $cos $\omega $t

Velocity in z-direction,

V_z = ${{dz} \over {dt}}$ = a$\omega $

Net velocity,

$\overrightarrow V $ = V_x$\widehat i$ + V_y$\widehat j$ + V_z$\widehat k$

Speed = $\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2} $

$ = \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}} $

$ = \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}} $

$ = \sqrt {2{a^2}{\omega ^2}} $

$ = \sqrt 2 a\omega $

Given,

$\overrightarrow v = K\left( {y\widehat i + x\widehat j} \right)$

$ \therefore $   Velocity in x direction,

${v_x} = {{dx} \over {dt}} = Ky\,$   . . . . . (1)

Velocity in y direction,

v_y = $\,{{dy} \over {dt}}$ = Kx . . . . . . . (2)

$ \therefore $   ${{\,{{dy} \over {dt}}} \over {{{dx} \over {dt}}}} = {{Kx} \over {Ky}}$

$ \Rightarrow $   ${{dy} \over {dx}} = {x \over y}$

$ \Rightarrow $   ydy $=$ xdx

Integrating both sides we get,

$\int {ydx} = \int {xdx} $

$ \Rightarrow $   ${{{y^2}} \over 2} = {{{x^2}} \over 2} + c$

$ \Rightarrow $   ${y^2} = {x^2} + 2c$

$ \therefore $   General equation,

y² = x² + constant.

Let the distance QM = l and distance RM = x.

Time to reach from Q to R is ${t_1} = {{l - x} \over v}$

Time to reach from R to P is ${t_2} = {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

Therefore, $t = {t_1} + {t_2} = {{l - x} \over v} + {{\sqrt {{x^2} + {d^2}} } \over {v/2}}$

On differentiating, we get

${{dt} \over {dx}} = {{0 - 1} \over v} + {1 \over {v/2}}{1 \over {2\sqrt {{x^2} + {d^2}} }} \times 2x$

$ \Rightarrow {{dt} \over {dx}} = {{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }}$

Put ${{dt} \over {dx}} = 0$, we get

${{ - 1} \over v} + {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = 0$

$ \Rightarrow {{2x} \over {v\sqrt {{x^2} + {d^2}} }} = {1 \over v}$

$ \Rightarrow 2x = \sqrt {{x^2} + {d^2}} \Rightarrow 4{x^2} = {x^2} + {d^2}$

$ \Rightarrow 3{x^2} = {d^2} \Rightarrow x = {d \over {\sqrt 3 }}$

Therefore, the distance $RM = x = {d \over {\sqrt 3 }}$, the time taken to reach P is minimum.

$\overrightarrow u = \widehat i + 2\widehat j = {u_x}\widehat i + {u_y}\widehat j$

$ \Rightarrow u\cos \theta = 1,u\sin \theta = 2$

Also $x = {u_x}t$ and

$y = {u_y}t - {1 \over 2}g{t^2}$

$ \Rightarrow $ $y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {u_x^2}}$

$\therefore$ $y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}$

We know, $R = {{{u^2}{{\sin }2}\theta } \over g}$ and $H = {{{u^2}{{\sin }^2}\theta } \over {2g}};$

${H_{\max }}\,\,$ is possible when $\theta = 90$$^\circ $

${H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2$

As $R = {{{u^2}\sin 2\theta } \over g}$

Range is maximum when projectile is thrown at an angle $45^\circ $.

$ \Rightarrow {R_{\max }} = {{{u^2}} \over g}$

${R_{\max }} = {{10 \times g \times 2} \over g} = 20$ meter

Maximum range of water coming out of fountain,

${R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}$

Total area around fountain,

$A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}$

$\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)$ ........(1)

Also $\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j$

$\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j$ ........(2)

Equating (1) and (2), we get

${{dx} \over {dt}} = ky\,\,\,\,\,\,$ .......(3)

and $\,\,\,\,\,{{dy} \over {dt}} = kx$ ......(4)

Dividing (3) and (4), we get

${{dy} \over {dx}} = {x \over y} $

$\Rightarrow ydy = xdx$

Integrating both sides of above equation, we get

$\int {ydy} = \int {xdx} $

$ \Rightarrow {y^2} = {x^2} + $ constant

Given $\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s$

$\overrightarrow v = \overrightarrow u + \overrightarrow a t $

$= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10$

$ = 7\widehat i + 7\widehat j$

We know speed is equal to magnitude of velocity.

$\therefore$ $\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,$ units

Average acceleration

$ = {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm int}} erval}}$

$ = {{\Delta \overrightarrow v } \over t}$

${\overrightarrow v _1} = +5\widehat i$, towards east direction.

$\overrightarrow {{v_2}} = +5\widehat j$, towards north direction

$\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}} $ = $5\widehat i - 5\widehat j$

$\therefore$ $\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}$

$\therefore$ $\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}$

$\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1$

$\therefore$ $\,\,\,\,\theta = {45^ \circ }$

Therefore the direction is North-west.

Range is same for angle of projection $\theta ,$ and ${90^ \circ } - \theta $

${T_1} = {{2u\sin \theta } \over g},\,\,{T_2} = {{2u\cos \theta } \over g}$

${T_1}{T_2} =$ ${{4{u^2}\sin \theta \cos \theta } \over {{g^2}}}$

= ${2 \over g} \times \left( {{{{u^2}\sin 2\theta } \over g}} \right)$

= ${{2R} \over g}$

(as $R = $${{{{u^2}\sin 2\theta } \over g}}$ )

Hence, ${T_1}{T_2}$ is proportional to $R.$

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

${{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,$

or $\cos \theta = {1 \over 2}$

$ \Rightarrow \cos \theta = \cos 60^\circ $

$ \Rightarrow \theta = 60^\circ $

From the figure it is clear that maximum horizontal range

$R = {{{u^2}\sin 2\theta } \over g}$

$ = {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}$

$ = 5\sqrt 3 $ = 8.66 m