Motion In A Plane Practice Questions

$y=5 x-5 x^2$ $y=x \tan \theta -\;\frac{1}{2} \;\frac{g x^2}{v^2}$ $\tan \theta =5=\;\frac{u_y}{u_x}$ $\Rightarrow u_y=5$

$\theta =45^{\circ}$ $R_1=\;\frac{R}{2}$ $\;\frac{u^2 \ sin 2 \theta _1}{g}=\;\frac{u^2 \ sin (90^{\circ}) }{2 g}$ $\Rightarrow 2 \theta _1=30^{\circ}$ $\theta _1=15^{\circ}$

$\Rightarrow v \cos \alpha =u \cos 45^{\circ} . . . . . \;\text{ (i) }\;$ $\& v \sin \alpha =u \sin 45^{\circ}-g t . . . . \;\text{ (ii) }\;$Solve for u we get$u=20 \sqrt{2} {\text{m}} / {\text{s}}$ $\Rightarrow H=\;\frac{u^2 \sin ^2 45^{\circ}}{20}=20 {\text{m}}$

$v_y=20 \times \sin \alpha -10 \times 10$ $v_x=20 \cos \alpha$ $\therefore \tan \beta =\;\frac{v_y}{v_x}=\;\frac{20 \sin \alpha -100}{20 \cos \alpha }$ $=\tan \alpha -5 \sec \alpha$

From graph,$v_{R G}=15 \sqrt{2} \tan 45^{\circ}$ $=15 \sqrt{2}$ $=\;\frac{30}{\sqrt{2}}$

$t=\;\frac{25 \sin \theta }{g}$and, $R=\;\frac{(25)^2(2 \sin \theta \cos \theta )}{g}$ $\Rightarrow R=\;\frac{25 \times 25 \times 2}{g} \times \;\frac{g t}{25} \times \cos \theta$ $\Rightarrow R=50 t \cos \theta$ $\therefore \tan \theta =\;\frac{g t}{25} \times \;\frac{50 t}{R}$ $=\;\frac{20 t^2}{R}$ $\Rightarrow \theta =\cot ^{-1}(\;\frac{R}{20 t^2})$

$\therefore u_1=u_2 sin \;\theta$ $\;\frac{H_1}{H_2}=\;\frac{\;\frac{u_1^2}{2 g}}{u_2^2 \;\frac{ sin \;^2 \theta }{2 g}}$ $= (\;\frac{u_1}{u_2 sin \;\theta }) ^2=1$

We know,Maximum height of a projectile $(H)=\;\frac{u^2 sin \;^2 \theta }{2 g}$Given, Ratio of initial velocity of two projectile$\;\frac{u_1}{u_2}=\;{{\sqrt{3}}}/{{\sqrt{2}}}$Both projectile reach the same maximum height.$\therefore H_1=H_2$ $\Rightarrow \;\frac{u_1^2 sin \;^2 \theta _1}{2 g}=\;\frac{u_2^2 sin \;^2 \theta _2}{2 g}$ $\Rightarrow u_1^2 sin \;^2 \theta _1=u_2^2 sin \;^2 \theta _2$ $\Rightarrow (\;\frac{u_1}{u_2}) ^2= (\;\frac{ sin \;\theta _2}{ sin \;\theta _1}) ^2$ $\Rightarrow (\;{{\sqrt{3}}}/{{\sqrt{2}}}) ^2= (\;\frac{ sin \;60^{\circ}}{ sin \;\theta _1}) ^2$ $\Rightarrow \;\frac{3}{2}=\;\frac{3}{4 sin \;^2 \theta _1}$ $\Rightarrow sin \;_{\theta _1}^2=\;\frac{1}{2}$ $\Rightarrow sin \;\theta _1=\;{1}/{{\sqrt{2}}}= sin \;45^{\circ}$ $\Rightarrow \theta _1=45^{\circ}$

$R=\;{ {\text{u}}^2 sin \;2 \theta }/{ {\text{g}}} {\text{R}}_{\max }=\;{ {\text{u}}^2}/{ {\text{g}}}=100$ ${\text{H}}_{\max }=\;{ {\text{u}}^2}/{2 {\text{g}}}=\;\frac{100}{2}=50 {\text{m}}$

${\text{At}} t=20 {\ s}$velocity of truck,$v=0+5 \times 20=100 {\text{m}} / {\ s}$At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck.Velocity of truck at $x$-direction $=100 {\text{m}} / {\ s}$ and in $y$-direction $=0$.$\therefore$ Velocity of ball $v_x=100 {\text{m}} / {\ s}, v_y=0$Now ball will show projectile motion where vertically downward acceleration ${\text{g}}=10 {\text{m}} / {\ s}$ act on the ball.As horizontally no acceleration acting on the ball so horizontal velocity $100 {\text{m}} / {\ s}$ will remain unchanged.Velocity of the ball when it reach the ground along y-direction after 1 sec.$v_y=0-10 \times 1$ $\Rightarrow v_y=-10 {\text{m}} / {\ s}$ $\therefore$ Velocity of ball $(\vec{v})=100 \hat{i}-10 \hat{j}$

The bomb is dropped from a fighter plane flying in horizontaldirection only. Let us assume the velocity of fighter plane, {v}_{{a}}=v_{x} {{i}}^{\^}. As fighter plane is moving in horizontal direction only. Hence, it has only one component. When fighter plane drops the bomb, the bomb will free fall from the fighter plane and have a horizontal velocity in the direction of plane and vertical velocity due to gravity. Velocity of bomb can be given as, {v}_{b}=v_{x} {{i}}^{\^ }-v_{y} {{j}}^{\^ }. Velocity in vertical direction will be negative as bomb is moving downwards. Now, for the person sitting inside the plane, hence the relative velocity of bomb with respect to plane can be calculated as {v}_{b a}={v}_{b}-{v}_{a}=v_{x} {{i}}^{\^}-v_{y} {{j}}^{\^}-v_{x} {{i}}^{\^}=-v_{y} {{j}}^{\^} The velocity of bomb with respect to person sitting in plane is only in vertical direction. Thus, the trajectory of bomb with respect to the person will be a straight line vertically down to the plane.

Given, velocity of mosquito, $v=(0.5 t^{2} \hat{i}+3 t \hat{j}+9 \hat{k}) {m} / {s}$ At $t=2 {s}, v=0.5(2)^{2} \hat{i}+3(2) \hat{j}+9 \hat{k}$ $=2 \hat{i}+6 \hat{j}+9 \hat{k}$ $\therefore$ Angle along $X$-axis, $\cos \alpha =\;{2}/{\sqrt{2^{2}+6^{2}+9^{2}}}=\;\frac{2}{\sqrt{121}}=\;\frac{2}{11}$ or $\alpha =\cos ^{-1}(\;\frac{2}{11})$ Similarly, angle along $Y$-axis, $\cos \beta =\;\frac{6}{11}$ $\Rightarrow \;\; \beta =\cos ^{-1}(\;\frac{6}{11})$ Angle along $Z$-axis, $\cos \gamma =\;\frac{9}{11}$ $\Rightarrow \;\; \gamma =\cos ^{-1}(\;\frac{9}{11})$ $\therefore$ Angle made by the direction of motion of mosquito will be $\cos ^{-1}(\;\frac{2}{11})$ from $X$-axis $=\tan ^{-1} \;\frac{\sqrt{117}}{2}$ $\cos ^{-1}(\;\frac{6}{11})$ from $Y$-axis $=\tan ^{-1} \;\frac{\sqrt{85}}{2}$ $\cos ^{-1}(\;\frac{9}{11})$ from $Z$-axis $=\tan ^{-1}(\;\frac{\sqrt{40}}{9})$

Given, equation of trajectory of projectile, $y=\alpha x-\beta x^{2}$ . . . (i) The standard equation of trajectory of projectile, $y=x \tan \theta -\;\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta }$ . . . (ii) Now, from Eqs. (i) and (ii), we get $\alpha =\tan \theta \;\text{ and }\; \beta =\;\frac{g}{2 u^{2} \cos ^{2} \theta }$ $\Rightarrow \;\; \alpha ^{2}=\;\frac{\sin ^{2} \theta }{\cos ^{2} \theta }$ $\therefore \;\; \;\frac{\beta }{\alpha ^{2}}=\;{g}/{2 u^{2} \cos ^{2} \theta \times \;\frac{\sin ^{2} \theta }{\cos ^{2} \theta }}=\;\frac{g}{2 u^{2} \sin ^{2} \theta } \times \;\frac{2}{2}$ $\;\frac{4 \beta }{\alpha ^{2}}=\;\frac{2 g}{u^{2} \sin ^{2} \theta }=\;\frac{1}{H} \Rightarrow H=\alpha ^{2} / 4 \beta$ $\;\;\; \theta =\tan ^{-1} \alpha \;\text{ and }\; H=\alpha ^{2} / 4 \beta$

Given, A person's swimming speed with respect to the river, $v_{m r}=10 \frac{m}{s}$ As shown in the figure, Let's determine the speed of the river flow, $v_{r}$ In X-direction, $v_{r}=v_{ {mr}} \sin 30^{\circ}$ $v_{r}=10 \times \sin 30^{\circ}$ $v_{r}=5 \frac{m}{s}$ Hence, the speed of the river flow is $5 \frac{m}{s}$. So, the value of the $x$ to the nearest integer is 5 .

Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them θ = 30°

{{V}}^{\to }_{{BW}}=4 \\sqrt{2} \cos 45 \hat{i}+4 \\sqrt{2} \sin 45 {{j}}^{\^} =4 \hat{i}+4 {{j}}^{\^} {{V}}^{ \to }_{{w}}=-{{j}}^{\^} ${\vec{V}}_{{B}}={{V}}^{\to }_{{BW}}+{\vec{V}}_{{w}}=4 \hat{i}+3 \hat{j}$ {\vec{S}}_{{B}}={\vec{V}}_{{B}} \\times {t}=(4 {{i}}^{\^}+3 \hat{j}) \\times 3 =12 \hat{i}+9 {{j}}^{\^} $|{\vec{S}}_{{B}}|=\\sqrt{(12)^{2}+(9)^{2}}=15 {m}$

Here, two projectiles are projected at angles 42° and 48° withsame initial velocity. As we know the expression of range of projectile, Range $=\frac{\text{u}^{2} \sin 2 \theta }{\text{g}}$ At $\theta _1$ = 42° Range, $\text{R}_{1}=\frac{\text{u}^{2} \sin 2(42)°}{\text{g}}=\frac{0.99 \text{u}^{2}}{\text{g}}$ At $\theta _2$ = 48° Range, $\text{R}_{2}=\frac{\text{u}^{2} \sin 2(48)°}{\text{g}}=\frac{0.99 \text{u}^{2}}{\text{g}}$ The range of the projectile is same for the two projectiles. Therefore, $\text{R}_1 = \text{R}_2$ Now, as we know the expression of height of the projectile, $\text{H}_{\max }=\frac{\text{u}^{2} \sin \theta }{2 \text{g}}$ At 42°, $\text{H}_{\max }=\text{H}_{1}=\frac{\text{u}^{2} \sin 42°}{2 \text{g}}=\frac{0.669 \text{u}^{2}}{2 \text{g}}$ At 48°, $\text{H}_{\max }=\text{H}_{2}=\frac{\text{u}^{2} \sin 48°}{2 \text{g}}=\frac{0.743 \text{u}^{2}}{2 \text{g}}$ Higher the value of q higher the value of maximum height.Therefore, $\text{H}_1 < \text{H}_2$.

Given, initial speed, $u=25 {ms}^{-1}$Angle of projection, $\theta =45^{\circ}$ Let, the maximum height $=H$Time taken to reach, ${H}= {t}$Acceleration due to gravity $g=10 {ms}^{-2}$Velocity at maximum height along $Y$-axis,$v_{y}=0 {ms}^{-1}$As we know that, for motion along$Y$-axis,$\because \;v_{y}^{2}-u_{y}^{2}=2 g H$ $\therefore \;0^{2}-(25 \sin 45^{\circ})^{2}=-2 \times 10 \times H$ $\Rightarrow -312.5=-20 H$ $\because \;H=15.625 {m}$ $\therefore \;H=\;\frac{1}{2} g t^{2}$ $\therefore \;=\sqrt{\;\frac{2 H}{g}}=\sqrt{\;\frac{2 \times 15.625}{10}}$ $\;=1.767=1.77 {s}$