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  • A projectile is given  an initial velocity of (i+2j)m/s where i is along the ground and j is along the vertical. If g=10m/s2, the equation of its trajectory is?<div class='qna-o

A projectile is given  an initial velocity of (i+2j)m/s where i is along the ground and j is along the vertical. If g=10m/s2, the equation of its trajectory is?

Option: 1

y=2x-5x^{2}
 


Option: 2

y=5x-2x^{2}


Option: 3

y=2x^{2}

 


Option: 4

y=3x-2x^{2}


Answers (1)

best_answer

 

Equation of path of a projectile -

     y= x \tan \theta\: - \: \frac{gx^{2}}{2u^{2}\cos ^{2}\theta }

It is equation of parabola, So the trajectory/path of the projectile is parabolic in nature.

                                                   g\rightarrow    Acceleration due to gravity

                                                   u\rightarrow  initial velocity

                                                   \theta =  Angle of projection

-

 

 

 

The equation of projectile is given by,

y=x\tan \theta-\frac{1}{2}\frac{gx^{2}}{u^{2}\cos ^{2}\theta}\\* \tan \theta=\frac{u_y}{u_x}= \frac{2}{1}= 2\\*

So cos\theta =\frac{1}{\sqrt{5}}

\therefore y= 2x-\frac{1}{2}\frac{10x^{2}}{\left ( 1^{2}+2^{2} \right )\times \frac{1}{5}}

\Rightarrow y= 2x-5x^{2}

Posted by

Rishabh

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