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the equation of projectile is y = ax - bx2 . its horizontal range is?

Given, y = ax - bx2    ........(1)
Since, equation of trajectory of projectile motion given angular projection is given by :
y = x tan θ - 12 gu2 cos2 θ x2      ......(2)Comparing eq. 1 with eq.2, we get :a = tan θor θ = tan-1 a   .....(3)and b = 12 gu2 cos2 θfrom above , we get u2 = g2 cos2 θ b    ........(4)
Since, horizontal range is given by :
 R = u2 sin 2 θ g  ....(5)Substituting the value of θ and u2 from eq. 3 and eq. 4 in eq. 5, we get :R = g2 b cos2 θ × sin 2  (tan-1 a)gR = sin 2 ( tan-1 a)2 b cos2 θ 
 

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Here you go !! Your answer :

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.)

CHEERS!!! :) (y)

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