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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
Given that y = ax - bx^2According 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 aii) Horizontal Range (R) = a / b (Ans.)
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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