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  • A body of mass  is being Projected with an angle of <img alt="30^{\circ}" src=

A body of mass 20 \mathrm{~kg} is being Projected with an angle of 30^{\circ} with the vertical. The trajectory of body is observed at point  (30,20) If the ' \mathrm{ T} ', is time of fight Then its momentum vector, at time \mathrm{ t=\frac{T}{\sqrt{3}}} is

Option: 1

\mathrm{ 20(38 \hat{\imath}+38(\sqrt{3}-2) \hat{\jmath}}


Option: 2

\mathrm{ 20(36 \hat{\imath}+30(\sqrt{3}-5) \hat{\jmath}}


Option: 3

\mathrm{ 10(36 \hat{\jmath}+30(\sqrt{3}-1) \hat{\jmath}}


Option: 4

None of these


Answers (1)

best_answer

\mathrm{ y=x \operatorname{tan} \theta-\frac{g x^2}{2 u^2 \cos ^2 \theta}}
\mathrm{ \begin{aligned} &\begin{aligned} \text { angle with Horizontal } & =90^{\circ}-30^{\circ} \\ & =60^{\circ} \end{aligned}\\ &\begin{aligned} y & =x \operatorname{tan} 60^{\circ}-\frac{1}{2} \times \frac{10 x^2}{4^2 \cos ^2 60^{\circ}} \\ & =x \sqrt{3}-\frac{2}{2} \times \frac{10 x^2}{u^2\left(\frac{1}{2}\right)^2} \\ & =\sqrt{3}-\frac{5 x^2}{4^2} \times 4 \\ y & =\sqrt{3} x-\frac{20 x^2}{u^2} \end{aligned} \end{aligned}}
\mathrm{ \begin{aligned} & \text { At }(30,20) \\ & 20=\sqrt{3} \times 30-\frac{20 \times 30 \times 30}{u^2} \\ & 2=3 \sqrt{3}-\frac{18000}{u^2} \\ & \frac{18000}{u^2}=\sqrt{3} \times 3-2 \\ & u^2=\frac{18000}{3 \sqrt{3}-2} \\ & u^2=\frac{18000}{3 \times 1.7-2}=5806.45 \approx 5806 \end{aligned}}\mathrm{ \begin{aligned} & =\frac{76 \sqrt{3}}{10}, \quad t=\frac{T}{\sqrt{3}} \\ & \vec{V}=u \cos \theta \hat{\imath}+(u \sin \theta-g t) \hat{\jmath} \\ & =76 \cos 60^{\circ} \hat{\imath}+\left(76 \sin 60^{\circ}-10 \times \frac{76}{10}\right) \hat{\jmath} \\ & =38 \hat{\imath}+(38 \sqrt{3}-76) \hat{\jmath} \\ & =30 \hat{\imath}+38 \sqrt{3} \hat{\jmath}-76 \hat{\jmath} \\ & \text { momentum }=m \vec{v} \\ & \end{aligned}}
\mathrm{ \begin{aligned} &u=76.2 \approx 76 \mathrm{~m} / \mathrm{sec}\\ &\begin{aligned} \text { Time } & =\frac{2 u \sin \theta}{y} \\ & =\frac{2 \times 76 \times \frac{53}{4}}{10 \times} \end{aligned} \end{aligned}}
\mathrm{ \begin{aligned} & =\frac{76 \sqrt{3}}{10}, \quad t=\frac{T}{\sqrt{3}} \\ & \vec{V}=u \cos \theta \hat{\imath}+(u \sin \theta-g t) \hat{\jmath} \\ & =76 \cos 60^{\circ} \hat{\imath}+\left(76 \sin 60^{\circ}-10 \times \frac{76}{10}\right) \hat{\jmath} \\ & =38 \hat{\imath}+(38 \sqrt{3}-76) \hat{\jmath} \\ & =30 \hat{\imath}+38 \sqrt{3} \hat{\jmath}-76 \hat{\jmath} \\ & \text { momentum }=m \vec{v} \\ & \end{aligned}}
\mathrm{ =20(38 \hat{i}+38 \sqrt{3} \hat{\jmath}-76 \hat{\jmath})}

Posted by

Divya Prakash Singh

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