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A particle is projected in the xy plane with the x axis along vertical, the point of projection in orign . The equation of the path is y = root3 – g x ^2 divided by two where y and x are in meters . Then the speed of projection in ms^-1 is a)2 B) root 3 c)4 d) root3 by 2


one year ago

Md Mushahid
21 Points
							A particle is projected in xy plane with x axis along vertical the point of projection is origin.the equation of the path is $y = \sqrt{3} x-\frac{g}{2} x ^{2}$ this is parabolic equation. hence question is based on projectile motion  if a particle is projected with speed u at an angle $\alpha$ to the vertical  equation of the path of the particle is given by$y =x\tan \alpha - \frac{g}{2u^{2}\times \cos ^{\alpha }}x^{2}$on camparing the both equation we get $\tan \alpha =\sqrt{3}$ and $2u^{2}\cos ^{2}\alpha =2$$\alpha = \arctan \sqrt{3} =60\degree$so,$\alpha = 60\degree$ $\Rightarrow$ $\cos \alpha = \cos 60\degree =\frac{1}{2}$ and now $2u^{2}\cos ^{2}\alpha =2$$u^{2}\times\left ( \frac{1}{2} \right )^{2} =1$$u^{2}=1\times 4$$u=\sqrt{4}$$u=2$Hence, speed of projection is 2m/s

one year ago
Khimraj
3007 Points
							Trajectory of projectile motion when u is intial speed inclined Ф angle with horizontalthe equation of projectile is : y = xTanФ - gx²/ 2u²Cos²Фand the given equation is : y = √3x - gx²/2on comparing both the equations ,we getTanФ = √3so, Ф = 60°HENCE, the angle of projectile is 60°and 2u²Cos²Ф = 2so u²Cos²Ф = 1u²cos²60° = 1u²*(1/2)² = 1u² = 2²so, u = 2

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions