Physics Motion in Plane

$h=\frac{1}{2}g{t}^2$

or $x=v\sqrt[]{\frac{2h}{g}}$

So, x = vt

or, $x=v\text{exception 25:}$

So, dist of man from helicopter is

$\sqrt[]{h^2+x^2}=\sqrt[]{h^2+v^2\frac{2h}{g}}=\sqrt[]{\frac{2v^{2}h}{g}+h^2}$

$a=\frac{v^2}{R}$

$\overrightarrow{a}=-\frac{v^2}{R}cos\theta \widehat{i}-\frac{v^2}{R}sin\theta \widehat{j}$

$R=\frac{\mathcal{l}}{\theta }$

$\Rightarrow Time=\frac{4*2\pi R}{V}=\frac{4*2\pi x}{V}\left(\frac{\mathcal{l}}{\theta }\right)$

Time $=\frac{4*2\pi *4.4*9.64*10^{15}}{8*1.5*10^{11}*\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3600$}\right.*\frac{\pi }{180}}$

$\Rightarrow Time=4.5*10^{10}socs$

$H_{max}=\frac{u^{2}si{n}^2\theta }{2g}=\frac{{\left(25*sin45°\right)}^2}{2*10}=15.625m$

$T=\frac{usin\theta }{g}=\frac{25*sin45}{10}=1.77s$

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

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

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

$h_1{h}_2=\frac{1}{4^2}{R}^2$

$R=4\sqrt[]{h_1{h}_2}$

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

Mass and temperature are scalars. We can fully describe them by magnitude alone and do not involve direction. 

Speed is a scalar quantity. It only has magnitude and no direction, unlike velocity, which is a vector.

Displacement, velocity, and acceleration are common vectors. These three vectors they have both magnitude and direction, and  follow vector addition rules.

Graphically, we place vectors head-to-tail. Analytically, we add components along axes using coordinate geometry, trigonometric functions, and unit vectors for precise vector resultant calculation.