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Grade: 11


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

Answers : (2)

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 byy =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

Hence, speed of projection is 2m/s

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