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

Answer: Range, [tex]R =\frac{a}{b}[/tex]Explanation:The equation of trajectory is:[tex]y = x tan \theta (1-\frac{x}{R})[/tex]Where, [tex]\theta[/tex] is the angle of projectile, R is the horizontal range.The equation of projectile is:y = ax-bx²[tex]\Rightarrow y = ax(1-\frac{b}{a}x)[/tex]On comparing:[tex]tan \theta = a[/tex][tex]R = \frac{a}{b}[/tex]Hence, the horizontal range is [tex]R =\frac{a}{b}[/tex]

y = a x - b x^2Range is a/by = tan Ф x -  g x²  / 2 u² cos² Фtan Ф = a        -  equation 1b = g / 2u² cos² Ф          so  u² cos² Ф = g /2b    - equation 2R = u cos Ф * 2 * u sin Ф / g  = 2/g  sinФ  u² cos Ф     = 2 /g   tan Ф   u² cos² Ф            by using equation 1  and equation 2     = (2 /g )  a  (g / 2b ) = a / b