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The trajectory of a projectile in a vert...

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, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.

A

The range of the projectile is ` a/b`

B

At ` x = a/(2b)` , the velocity of projectile becomes zero

C

The maximum height attained by projectile is ` a^(2)/(4b)`

D

The angle of projection is ` tan^(-1)(a)`

Text Solution

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The correct Answer is:
A, C, D
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