Physics Motion in Plane

The graphical method using the head-to-tail or parallelogram laws only helps in visualising vectors and their resultants. But, it has limited accuracy. Because they cannot be precise when you consider the scale and angles. That's why it is important to use vector addition using the analytical method. That involves combining vector components. The graphical approach is primarily for conceptual understanding.

Scalar quantities only have magnitude, which makes sense to combine using ordinary algebra. But vector quantities have both magnitude and direction. Due to this directional aspect, vectors must obey special rules of vector algebra. Vectors have to specifically follow the triangle law or the parallelogram law of addition to be represented in the graph format. These graphical methods account for both magnitude and direction. This makes sure that the resultant vector accurately reflects the combined effect of the individual vectors. If we apply ordinary algebra, we won't be able to know the directional information.

First angle, θ₁ = θ, $R=\frac{u^{2}sin2\theta }{g}$   

Another angle (θ₂ = 90 - θ) for which range will be

Same as that of θ₁ = θ

$R^1=\frac{u^{2}sin2\left(90-\theta \right)}{g}=\frac{u^2}{g}sin2\theta =R$

at  ${\theta }_1=\theta ,h_1=\frac{u^{2}si{n}^2\theta }{2g}$

$\&{\theta }_2=90-\theta ,h_2=\frac{u^{2}si{n}^2{\theta }_2}{2g}=\frac{u^{2}co{s}^2\theta }{2g}$

$\Rightarrow h_1{h}_2=\frac{1}{4^2}{\left[\frac{u^{2}sin2\theta }{g}\right]}^2$  

$R=4\sqrt[]{h_1{h}_2}$  

So, Both statement is true & Reason is correct explanation for statement 1.

We know that horizontal speed will remain constant

20 cos = v cos ……… (i)

Along y-axis

$U_y=20sin\alpha ,V_y=vsin\beta$

$V_y=V_y+a_{y}t$

$$ $Vsin\beta =20sin\alpha -10*10$  ……. (ii)

$\Rightarrow (ii)÷(i)$

$\Rightarrow tan\beta =tan\alpha -5sec\alpha$

d = R

$60=R\left[\frac{3\pi }{4}\right]$

$R=\frac{60*4}{3\pi }=\frac{80}{\pi }m$

Displacement = $\sqrt[]{R^2+R^2-2R^{2}cos135}$ $\text{exception 25:}$

= $\sqrt[]{2R^2-2R^2\left(-0.7\right)}$ $\text{exception 25:}$

= $\sqrt[]{3.4R^2}$ $\text{exception 25:}$

= $\sqrt[]{3.4{\left(\frac{80}{4}\right)}^2}$ $\text{exception 25:}$

47 m

$H=\frac{u^{2}si{n}^2\theta }{2g}andR=\frac{u^{2}sin2\theta }{g}$

According to question

$R=H\Rightarrow \frac{u^{2}si{n}^2\theta }{2g}=\frac{2u^{2}sn\theta cos\theta }{g}\Rightarrow tan\theta =4$

$\mathrm{}|\stackrel{?}{\mathrm{P}}+\stackrel{?}{\mathrm{Q}}|=\left|\stackrel{?}{\mathrm{P}}\right|$

^{${\mathrm{P}}^2+{\mathrm{Q}}^2+2\mathrm{P}\mathrm{Q}\mathrm{c}\mathrm{o}\mathrm{s}?\theta ={\mathrm{P}}^2$}

$\Rightarrow \mathrm{}Q+2P\mathrm{c}\mathrm{o}\mathrm{s}?\theta =0$

$\Rightarrow \mathrm{}\mathrm{c}\mathrm{o}\mathrm{s}?\theta =-\frac{\mathrm{Q}}{2\mathrm{P}}$

$\mathrm{t}\mathrm{a}\mathrm{n}?\alpha =\frac{2\mathrm{P}\mathrm{s}\mathrm{i}\mathrm{n}?\theta }{2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}}=\mathrm{\infty }$

$?\mathrm{}[2\mathrm{P}\mathrm{c}\mathrm{o}\mathrm{s}?\theta +\mathrm{Q}=0]$

$\alpha ={90}^{?}$

From conservation of Energy

$\frac{1}{2}m{v}_0^2=mgh$

$v_0=10\sqrt[]{2}$

For A to B

$a=-gsin45°=\frac{-10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l}=2\sqrt[]{2}+2\\ =2\left(\sqrt[]{2}+1\right)\end{array}$

 $T_1=\frac{2u}{g}{T}_2=\frac{2Vsin\theta }{g}$

$As,T_1=T_2\Rightarrow \frac{2u}{g}=\frac{2Vsin\theta }{g}\Rightarrow u=vsin\theta$ - (i)

$\frac{H_1}{H_2}=\frac{u^2}{2g}*\frac{2g}{v^{2}si{n}^2\theta }={\left(\frac{u}{vsin\theta }\right)}^2=1$

after 2 sec, v = 20 m/sec

$\begin{array}{l}\overrightarrow{v}=ucos45°\widehat{i}+\left(usin45°-gt\right)\widehat{j}\\ v=20=\sqrt[]{{\left(ucos45°\right)}^2+{\left(usin45°-20\right)}^2}\end{array}$

$\Rightarrow 400=u^2+400-40usin45°$

u = 40 sin 45°

$H_{max}=\frac{u^{2}si{n}^{2}45}{2g}=\frac{40^{2}si{n}^{2}45°*si{n}^{2}45}{2g}$

$=\frac{40*40}{2*10}*\frac{1}{2}*\frac{1}{2}=20m$