Physics, asked by narendra9584, 11 months ago

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

Answers

Answered by Sukhpreet85
4

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

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