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The equation of projectile is y=ax-bx×bx. It`s horizontal range is

The equation of projectile is y=ax-bx×bx. It`s horizontal range is

Grade:12th pass

1 Answers

Arun
25757 Points
5 years ago

Given that y = ax - bx^2

According to Equation of Trajectory, y = x Tan (theta) [ 1 - x / R ]

y = ax - bx^2 can also be written as,

= y = xa - bx^2

= y = xa [ 1 - bx / a]

= y = xa [ 1 - x / a / b ]

Now comparing this equation with Equation of trajectory, we get,

i) Tan (theta) = a = theta = Tan^-1 a

ii) Horizontal Range (R) = a / b (Ans.)

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