Question

The equation of motion of projectile is y is equals to a x minus 3 x square where a and b are the constant of motion the horizontal range of the projectile is

Answer 1

(1)

Arun

20942 Points

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.)