Motion In A Plane Practice Questions

$\vec{v}=k(y {i}^{∧}+x {j}^{∧})$.....(1) Also $\vec{v}=v_x {i}^{∧}+v_y {j}^{∧}$ $\vec{v}=\;\frac{d x}{d t} {i}^{∧}+\;\frac{d y}{d t} {j}^{∧}$ .....(2)Equating (1) and (2), we get$\;\frac{d x}{d t}=k y$......(3)and $\;\frac{d y}{d t}=k x$.....(4) Dividing (3) and (4), we get$\;\;\frac{d y}{d x}=\;\frac{x}{y}$ $\;\Rightarrow y d y=x \text{dx}$ Integrating both sides of above equation, we get$\;∫ y d y=∫ x \text{dx}$ $\;\Rightarrow y^2=x^{2+}\;\text{ constant }\;$

$\;\;\text{ Given }\; \vec{u}=3 {i}^{∧}+4 {j}^{∧}, \vec{a}=0.4 {i}^{∧}+0.3 {j}^{∧}, t=10 s$ $\;\vec{v}=\vec{u}+\vec{a} t$ $\;=3 {i}^{∧}+4 {j}^{∧}+(0.4 {i}^{∧}+0.3 {j}^{∧}) \times 10$ $\;=7 {i}^{∧}+7 {j}^{∧}$We know speed is equal to magnitude of velocity.$\therefore |\vec{v}|={\sqrt{7^{2+}7^2}}=7 {\sqrt{2}}$ units

Average acceleration $\;=\;\frac{\;\text{ change \in velocioty }\;}{\;\text{ time interval }\;}$ $\;=\;\frac{\Delta \vec{v}}{t}$ $\vec{v_1}=+5 {i}^{∧}$, towards east direction$\vec{v_2}=+5 {j}^{∧}$, towards north direction $\;\Delta \vec{v}=\vec{v_2}-\vec{v_1}=5 {i}^{∧}-5 {j}^{∧}$ $\; \therefore \vec{a}=\;\frac{5 {j}^{∧}-5 {i}^{∧}}{10}=\;\frac{{j}^{∧}-{i}^{∧}}{2}$ $\; \therefore \;\; a=\;{{\sqrt{1^{2+}(-1)^2}}}/{2}=\;{{\sqrt{2}}}/{2}=\;{1}/{{\sqrt{2}}} m s^{-2}$ $\;\tan \theta =\;\frac{v_2}{v_1}=\;\frac{5}{5}=1$ $\; \therefore \;\; \theta =45^{\circ}$Therefore the direction is North-west.

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,$\;\frac{v_0}{2}=v_0 \cos \theta$or $\cos \theta =\;\frac{1}{2}$ $\Rightarrow \cos \theta =\cos 60^{\circ}$ $\Rightarrow \theta =60^{\circ}$

Range is same for angle of projection $\theta$, and $90^{\circ}-\theta$ $\;T_1=\;\frac{2 u sin \;\theta }{g}, T_2=\;\frac{2 u \cos \theta }{g}$ $\;T_1 T_2=\;\frac{4 u^2 sin \;\theta \cos \theta }{g^2}$ $\;=\;\frac{2}{g} \times (\;\frac{u^2 sin \;2 \theta }{g})$ $\;=\;\frac{2 R}{g}$ $\;\;\text{ (as }\; R=\;\frac{u^2 sin \;2 \theta }{g} \;\text{ ) }\;$Hence, $T_1 T_2$ is proportional to $R$.

From the figure it is clear that maximum horizontal range $R=\;\frac{u^2 sin \;2 \theta }{g}$ $=\;\frac{(10)^2 sin \; (2 \times 30^{\circ}) }{10}$ $=5 {\sqrt{3}}=8.66 {\text{m}}$