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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 isa)2                                          B)root 3                                    c)4                                           d)root3 by 2

Md Mushahid
21 Points
one year ago
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

Khimraj
3007 Points
one year ago
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