Motion In A Plane Practice Questions

$\; {\text{H}}_{\max }=\;{ {\text{u}}^2 sin \;^2 \theta }/{2 {\text{g}}}$ $\;\;{ {\text{H}}_{1 \max }}/{ {\text{H}}_{2 \max }}=\;{ {\text{u}}_1^2}/{ {\text{u}}_2^2}$ $\;\;{64}/{ {\text{H}}_{2 \max }}=\;{ {\text{u}}^2}/{( {\text{u}} / 2)^2}$ $\; {\text{H}}_{2 \max }=16 {\text{m}}$

Time of flight $=5+3=8 {\ sec}$.Now, $\;\; T=\;\frac{2 u sin \;\theta }{g}$ $\;8=\;{2 {\text{u}} \cdot sin \;30^{\circ}}/{10}$ $\;\Rightarrow {\text{u}}=80 {\text{m}} / {\ sec}$Ans.

$\;\;\text{ So }\; {\text{R}}=\;{ {\text{u}}^2 sin \;(2 \times 15)}/{ {\text{g}}}=\;{ {\text{u}}^2}/{2 {\text{g}}}=\;\text{ So }\; \Rightarrow \;{ {\text{u}}^2}/{ {\text{g}}}=100$ $\; {\text{R}}^{'}=\;{ {\text{u}}^2 sin \;(2 \times 45)}/{ {\text{g}}}=\;{ {\text{u}}^2}/{ {\text{g}}}=100 {\text{m}}$

$\; {\text{R}}=\;{ {\text{u}}^2 sin \;2 \theta }/{ {\text{g}}}$ $\; \{ {\text{u}}_1=40 {\text{m}} / {\ s}, \theta _1=30^{\circ}, {\text{u}}_2=60 {\text{m}} / {\ s}, \theta _2=60^{\circ}\}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}= (\;{ {\text{u}}_1}/{ {\text{u}}_2}) ^2 \;\frac{ sin \;2 \theta _1}{ sin \;2 \theta _2}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}= (\;\frac{40}{60}) ^2 \times \;\frac{ sin \;60^{\circ}}{ sin \;120^{\circ}} \Rightarrow \;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{4}{9}$

time to cross the River width $\omega =1000 {\text{m}}$ $\;\text{ is }\;=\;{1 {\text{km}}}/{4 {\text{km}} / {\text{h}}}$Drift ${\text{x}}= {\text{Vm}} / {\text{g}} \times {\text{t}}$Where ${\text{Vm}} / {\text{g}}$ is velocity of River w.r. to ground. $\; {\text{x}}= {\text{Vm}} / {\text{g}} \times \;\frac{1}{4}=750 {\text{m}}=\;\frac{3}{4} {\text{km}}$ $\; {\text{Vm}} / {\text{g}}=3 {\text{km}} / {\text{hr}}$Ans is $3 {\text{km}} / {\text{hr}}$.

$\; {\text{v}}_{ {\text{y}}}={\sqrt{2 g h}}={\sqrt{200}}$ $\;v_{\;\text{net }\;}={\sqrt{25+200}}=15 {\text{m}} / {\ s}$

$\;y=x-\;\frac{x^2}{20}$ $\; (\;\frac{d y}{d x}) =1-\;\frac{x}{10} \;\text{ for }\; y_{\max } ; \;\frac{d y}{d x}=0$ $\;x=10$ $\;y_{\max }=10-\;\frac{100}{20}=5 m$

In projectile motion, ${\text{H}}_{\;\text{many }\;}=\;{ {\text{u}}^2 sin \;^2 \theta }/{2 {\text{g}}}$ at $\theta =30^{\circ}, \;\; {\text{H}}_1=\;{ {\text{u}}^2 sin \;^2 30^{\circ}}/{2 {\text{g}}}$ at $\theta =60^{\circ}, \;\; {\text{H}}_2=\;{ {\text{u}}^2 sin \;^2 60^{\circ}}/{2 {\text{g}}}$ $\;{H_1}/{ {\text{H}}_2}=\;\frac{ sin \;^2 30^{\circ}}{ sin \;^2 60^{\circ}}=\;{ (\frac{1}{2}) }/{ (\;{{\sqrt{3}}}/{2}) ^2}=\;\frac{1}{3}$

$\;\text{ Range }\;=\;{ {\text{u}}^2 sin \;2 \theta }/{ {\text{g}}}$Range for projection angle " $\alpha$ "${\text{R}}_{ {\text{l}}}=\;{ {\text{u}}^2 sin \;2 \alpha }/{ {\text{g}}}$Range for projection angle " $\beta$ "$\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 \beta }/{ {\text{g}}}$ $\;\alpha +\beta =90^{\circ} \;\text{ (Given) }\;$ $\;\Rightarrow \beta =90^{\circ}-\alpha$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 (90^{\circ}-\alpha ) }/{ {\text{g}}}$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \; (180^{\circ}-2 \alpha ) }/{ {\text{g}}}$ $\; {\text{R}}_2=\;{ {\text{u}}^2 sin \;2 \alpha }/{ {\text{g}}}$ $\;\Rightarrow \;{ {\text{R}}_1}/{ {\text{R}}_2}=\;{ ({ {\text{u}}^2 sin \;2 \alpha }/{ {\text{g}}}) }/{ (\;{ {\text{u}}^2 sin \;2 \alpha }/{ {\text{g}}}) }=\;\frac{1}{1}$ $\;$

For a ground to ground projectile, Horizontal range is given by $R=\;\frac{u^2 sin \;2 \theta }{g}$ And for $R_{\max }, sin \;2 \theta$ must be maximum.

Since range is same.$\Rightarrow \theta _1+\theta _2=90^{\circ}$ $\Rightarrow \theta _2=30^{\circ}$ $\Rightarrow (H_{\max }) _1+ (H_{\max }) _2=\;\frac{u^2 sin \;^2 \theta _1}{2 g}+\;\frac{u^2 sin \;^2 \theta _2}{2 g}$ $=\;\frac{40^2}{20} (\;\frac{1}{4}+\;\frac{3}{4}) =80 {\text{m}}$

$\; {\text{u}} \cos \theta =\;{{\sqrt{3}} {\text{u}}}/{2} \Rightarrow \cos \theta =\;{{\sqrt{3}}}/{2}$ $\;\Rightarrow \theta =30^{\circ}$ $\; {\text{T}}=\;{2 {\text{u}} sin \;30^{\circ}}/{ {\text{g}}}=\;{ {\text{u}}}/{ {\text{g}}}$Option 2.

$\; {\text{U}}_{ {\text{x}}}=40 \cos 30=20 {\sqrt{3}}$ $\; {\text{U}}_{ {\text{y}}}=40 sin \;30=20$ $\; {\text{Vx}}=20 {\sqrt{3}}$ $\; {\text{V}}_{ {\text{y}}}= {\text{u}}_{ {\text{y}}}- {\text{gt}}=20-10 \times 2=0$ $\; {\text{V}}=\sqrt{ {\text{v}}_{ {\text{x}}}{ }^{2+} {\text{v}}_{ {\text{y}}}{ }^2}=20 {\sqrt{3}}= {\text{ms}}^{-1}$

$\; {\text{H}}_{\max }=\;{ {\text{v}}^2}/{2 {\text{g}}}=136 {\text{m}}$ $\; {\text{R}}_{\max }=\;{ {\text{v}}^2}/{ {\text{g}}}=2 {\text{H}}_{\;\text{max }\;}$ $\;=2(136)$ $\;=272 {\text{m}}$

To hit the jet both should have same horizontal component of velocity.To hit the jet$400 \cos \theta =200$ $\Rightarrow \cos \theta =\;\frac{1}{2}$ $\Rightarrow \theta =60^{\circ}$

Let projection speed is $u$ ${\text{R}}_1=\;{ {\text{u}}^2 {\text{Sin}}(90^{\circ})}/{g} ; {\text{R}}_2=\;{ {\text{u}}^2 \sin (60^{\circ})}/{g}$ $\;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{2}{\sqrt{3}}$

$x=3 t \Rightarrow a_x=0$ $y=5 t^3 \Rightarrow a_y=30 t$ $\vec{a}(t=1)=30 \hat{y}$

$h_1=\;\frac{u^2 \sin ^2 \theta }{2 g}$ $h_2=\;\frac{u^2 \cos ^2 \theta }{2 g}$ $\therefore \sqrt{h_1 h_2}=\;\frac{u^2 \sin \theta \cos \theta }{2 g}$ $=\;\frac{R}{4}$ $\Rightarrow R=4 \sqrt{h_1 h_2}$