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The equation of trajectory of a projectile is given by y=x -\frac{x^{2}}{100} Find the maximum height

Option: 1

100 m


Option: 2

25 m


Option: 3

300 m


Option: 4

2 m


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

-

 

 

 

 Comparing=x-\frac{x^{2}}{100}

With the equation y= \tan \theta \times x-\frac{1}{2}\frac{g}{u^{2}\cos ^{2}\theta }x^{2}

We get the angle of projection to be \theta = 45

Maximum height is given by

H=\frac{v_{0}^{2}\sin ^{2}\theta }{2(10))}

=25 m

Hense 2 is correct

Posted by

Pankaj Sanodiya

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