Question

The equation of projectile is y=ax-bx^2 then horizontal range is

Answer 1

Y = a x - b x^2

Range is a/b

y = tan Ф x - g x² / 2 u² cos² Ф
tan Ф = a - equation 1
b = g / 2u² cos² Ф so u² cos² Ф = g /2b - equation 2

R = 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