Physics - Kinematics Question with Solution | TestHub

Here, particle is moving in two - dimensional (x – y) plane, with constant speed so slope of

graph$\frac{\text{d}\text{y}}{\text{d}\text{x}}=\frac{{\text{v}}_{\text{y}}}{{\text{v}}_{\text{x}}}$and${\text{v}}_0=\sqrt{{\text{v}}_{\text{x}}^2+{\text{v}}_{\text{y}}^2}\text{.}$

As slope,$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{3}}=\frac{{\text{v}}_{\text{y}}}{{\text{v}}_{\text{x}}}$

So,${\text{v}}_{\text{x}}=-{\text{v}}_{\text{y}}\sqrt{3}$

On squaring and adding

${\text{v}}_{\text{x}}^2+{\text{v}}_{\text{y}}^2={\text{v}}_0^2$

${3\text{v}}_{\text{y}}^2+{\text{v}}_{\text{y}}^2={\text{v}}_0^2$

${\text{v}}_{\text{y}}=\pm \frac{{\text{v}}_0}{2}$

${\text{v}}_{\text{x}}=\mp \frac{\sqrt{3}{\text{v}}_0}{2}$

So,$\text{y}=\alpha {\text{e}}^{\frac{-2}{\sqrt{3}{\text{v}}_0}\times \frac{\sqrt{3}{\text{v}}_0}{2}}$

$\text{y}=\alpha {\text{e}}^{-1}$