Motion In A Plane Practice Questions

In projectile motion, the horizontal component of velocity remains constant throughout the flight, so $u_x = v_x$.
Given the initial angle $\theta_0 = 60^{\circ}$ and the final angle $\theta = 45^{\circ}$ where the speed is $v = 20 \text{ m/s}$, we set the horizontal components equal:
$u \cos(60^{\circ}) = v \cos(45^{\circ})$
Substituting the known values:
$u \cdot \frac{1}{2} = 20 \cdot \frac{1}{\sqrt{2}}$
Solving for $u$:
$u = 2 \cdot \frac{20}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ m/s}$

To minimize the time taken to cross the river, the boat must steer perpendicular to the river bank.

1. Convert speeds to SI units: $v_r = 18 \times \frac{5}{18} = 5 \text{ m/s}$ and $v_b = 36 \times \frac{5}{18} = 10 \text{ m/s}$.
2. Time for one leg of the journey: $t = \frac{d}{v_b} = \frac{200 \text{ m}}{10 \text{ m/s}} = 20 \text{ s}$.
3. Total time for a round trip: $T = 2t = 2 \times 20 = 40 \text{ s}$.
4. Drift per leg due to river current: $x = v_r \times t = 5 \text{ m/s} \times 20 \text{ s} = 100 \text{ m}$.
5. Total displacement along the river bank after both legs: $X_{total} = 100 \text{ m} + 100 \text{ m} = 200 \text{ m}$.

Displacement $=2 {\sqrt{2}} {\text{km}}$

At height pointThe speed is $v_x=v_{\cos } 45^{\circ}$ $v_x=\;{v_0}/{{\sqrt{2}}}$So angular momentum about origin is$L=m v_x \cdot H$ $\;\Rightarrow L=m \;{v_0}/{{\sqrt{2}}} \cdot (\;{v_0}/{{\sqrt{2}}}) ^2 \times \;\frac{1}{2 g}$ $\;\Rightarrow L=\;{m v_0^3}/{4 {\sqrt{2}} g}(-\mu -z \;\text{-axis }\;)$

$H_{\max }\;=\;\frac{u^2 sin \;^2 \theta }{2 g}$ $\;\frac{H_1}{H_2}\;=\;\frac{ sin \;^2(45-\alpha )}{ sin \;^2(45+\alpha )}$ $\;=\;\frac{(\cos \alpha - sin \;\alpha )^2}{(\cos \alpha + sin \;\alpha )^2}$ $\;\frac{H_1}{H_2}\;=\;\frac{1- sin \;2 \alpha }{1+ sin \;2 \alpha }$

$\;R=3 H$ $\;\;\frac{u^2 sin \;2 \theta }{g}=\;\frac{3 u^2 sin \;^2 \theta }{2 g}$ $\;2 sin \;\theta \cos \theta =\;\frac{3}{2} sin \;\theta sin \;\theta$ $\;\;\frac{4}{3}=\tan \theta$ $\;\theta =53^{\circ}$ $\;R=\;\frac{u^2}{g} \times 2 \times \;\frac{4}{5} \times \;\frac{3}{5}=\;\frac{24}{25} \;\frac{u^2}{g}$

At highest point $v=u \cos \theta =10 {\text{m}} / {\ s}$ $\;k_1=\;\frac{1}{2} m v^2, k_2=\;\frac{1}{2} m v^2$

Time of flight for $1^{\;\text{st }\;}$ projectile$T_1=\;\frac{24 sin \;(45+\alpha )}{g}$And $T_2=\;\frac{24 sin \;(45-\alpha )}{g}$ $\;\;\text{ So, }\; \;\frac{T_1}{T_2}=\;\frac{ sin \;(45+\alpha )}{ sin \;(45-\alpha )}=\;{\;{\cos \alpha }/{{\sqrt{2}}}+{ sin \;\alpha }/{{\sqrt{2}}}}/{{\cos \alpha }/{{\sqrt{2}}}-\;{ sin \;\alpha }/{{\sqrt{2}}}}$ $\;\Rightarrow \;\frac{T_1}{T_2}=\;\frac{1+\tan \alpha }{1-\tan \alpha }$

$\;U=60 m / s$ $\;\theta =30^{\circ}$ $\;t_{H_{\max }}=\;\frac{U sin \;\theta }{g}=\;\frac{60}{10} \times \;\frac{1}{2}=3 s$ $\;\;\text{ Height \in }\; 1^{\;\text{st }\;} \;\text{ second }\;=30 \times 1-\;\frac{1}{2} \times 10 \times 1$ $\;=25 m$ $\;\;\text{ Height \in last second }\;=5 m$ $\;\;\frac{25}{5}=5$

$t\;=\;\frac{w}{v_b \cos (60)}$ $w\;=t v_b \cos (60)$ $\;=30 \times 27 \times \;\frac{5}{18} \times \;\frac{1}{2}$ $\;=112.5 {\text{m}}$

$\;\vec{r}= (5 t^2 {i}^{∧}-5 t {i}^{∧})$ $\;\;\text{ So }\; \;\frac{d \vec{r}}{d t}=\vec{v}=(10 t {i}^{∧}-5 {j}^{∧})$ $\;\;\text{ At }\;(t=2 \;\text{ second }\;)$ $\; \;\;\;\; \vec{v}(t=2)=(20 {i}^{∧}-5 {j}^{∧})$ $\;|\vec{v}|=5 {\sqrt{1+16}}=5 {\sqrt{17}} m / s$

$h=\;\frac{u^2 sin \;^2 \theta }{2 g}\;\theta =90^{\circ}-\varphi$ $\Rightarrow h=\;\frac{u^2 \cos ^2 \theta }{2 g}\;$From $\varphi =0$ to $\varphi =90$ $h$ decreases with $\cos ^2$ function.

$H_1=\;\frac{u^2 sin \;^2 \theta _1}{2 g}\;\;\;H_2=\;\frac{u^2 sin \;^2 \theta _2}{2 g}$ $T_1=\;\frac{2 u sin \;\theta _1}{g}\;\;\;T_2=\;\frac{2 u sin \;\theta _2}{g}$ $\;\frac{H_1}{H_2}=\;\frac{ sin \;^2 \theta _1}{ sin \;^2 \theta _2}\;\;\;\;\frac{T_1}{T_2}=\;\frac{ sin \;\theta _1}{ sin \;\theta _2}$ $8= (\;\frac{T_1}{T_2}) ^2\;\Rightarrow \;\frac{T_1}{T_2}=2 {\sqrt{2}}$ $T_1: T_2=2 {\sqrt{2}}: 1\;$

$v_y=10 m / s$ $v_x=v_{\;\text{river }\;}+v_{\;\text{boat }\;}\;=27+9$ $\;=36 km / h=10 m / s$ $R= (\;\frac{2 v_y}{g}) v_x\;=\;\frac{2 \times 10 \times 10}{10}=20 m$ $\;=2000 cm$

$\;x=2+4 t$ $\;\;\frac{d x}{d t}=v_x=4$ $\;\;\frac{d v_x}{d t}=a_x=0$ $\;y=3 t+8 t^2$ $\;\;\frac{d y}{d t}=v_y=3+16 t$ $\;\;\frac{d v_y}{d t}=a_y=16$the motion will be uniformly accelerated motion. For path$\; {\text{x}}=2+4 {\text{t}}$ $\;\;{( {\text{x}}-2)}/{4}= {\text{t}}$Put this value of $t$ is equation of $y$ $y=3 (\;\frac{x-2}{4}) +8 (\;\frac{x-2}{4}) ^2$this is a quadratic equation so path will be parabola.Correct answer (4)

{ {\text{v}}}^{\to }= \vec{ds} / {\text{dt}} =4 {\text{t}} {\text{j}}^{\^} At {\text{t}}=1 \ sec { {\text{v}}}^{\to }=4 { {\text{j}}^{\^}}

The ball needs to just cross 4 steps to just hit $5^{\;\text{th }\;}$ stepTherefore, horizontal range $( {\text{R}})=0.4 {\text{m}}$ ${\text{R}}=\;\text{ u.t }\;$Similarly, in vertical direction$\; {\text{h}}=\;\frac{1}{2} {\text{gt}}^2$ $\;0.4=\;\frac{1}{2} {\text{gt}}^2$ $\;0.4=\;\frac{1}{2} {\text{g}} (\;{0.4}/{ {\text{u}}}) ^2$ $\; {\text{u}}^2=2$ $\; {\text{u}}={\sqrt{2}} {\text{m}} / {\ s}$Therefore, ${\text{x}}=2$

$\;u_x=5 {\text{m}} / {\ s} \;\; {\text{a}}_{ {\text{x}}}=3 {\text{m}} / {\ s}^2 \;\; {\text{x}}=84 {\text{m}}$ $\; {\text{v}}_{ {\text{x}}}^{2-} {\text{u}}_{ {\text{x}}}^2=2 {\text{ax}}$ $\; {\text{v}}_{ {\text{x}}}^{2-}25=2(3)(84)$ $\; {\text{V}}_{ {\text{x}}}=23 {\text{m}} / {\ s}$ $\; {\text{v}}_{ {\text{x}}}- {\text{u}}_{ {\text{x}}}= {\text{a}}_{ {\text{x}}} {\text{t}}$ $\; {\text{t}}=\;\frac{23-5}{3}=6 {\ s}$ $\; {\text{v}}_{ {\text{y}}}=0+ {\text{a}}_{ {\text{y}}} {\text{t}}=0+2 \times (6)=12 {\text{m}} / {\ s}$ $\; {\text{v}}^2= {\text{v}}_{ {\text{x}}}^{2+} {\text{v}}_{ {\text{y}}}^2=23^{2+}12^2=673$ $\; {\text{v}}={\sqrt{673}} {\text{m}} / {\ s}$

$\;\;\frac{u^2 sin \;2 \theta }{g}=\;\frac{u^2 sin \;^2 \theta }{2 g}$ $\;4 sin \;\theta \cos \theta = sin \;^2 \theta$ $\;4=\tan \theta$