Question

6. Equation of motion of projectile in x-y plane is
y = ax - bx^2, then the angle of projection with
x-axis is
[NCERT Pg. 78]
(1) tan^-1 (b)
(2) tan^-1(a)
(3) cos^-1(b)
(4) sin^-1(a)​

Answer 1

Answer:

The range of the projectile is ab

At x=a2b , the velocity of projectile becomes zero

The maximum height attained by projectile is a24b

The angle of projection is tan−1(a)

Answer :

A::C::D

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Explanation: