Question

the trajectory of a projectile in a vertical plane is y=ax-bx^2 where a and b are constant and X and y are horizontal and vertical distance of projectile from the point of projection.the maximum height and angle of projection from horizontal are​

Answer 1

Explanation:

The trajectory of a projectile in a vertical plane is y=ax-bx2,where a and b are constants,and x and y are respectievly the horizontal and the verticaldistanceof the projectile from the point of projection.The maximum height attained is?and the angle of projectionfrom the horizontal is?

Asked by Aditi Priya1st October 2010, 8:33 AM

Answer 2

Answer:

Explanation:

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