Equation of trajectory of a projectile is given by `y = -x^(2) + 10x ` where `x` and `y` are in meters and `x` is along horizontal and `y` is vericall `y` upward and particle is projeted from origin. Then : `(g = 10) m//s^(2)`

Options

A. initial velocity of particle is `sqrt(505) m//s`
B. horizontal range is `10 m`
C. maximum height is `25 m`
D. angle of projection with horizontal is `tan^(-1)(5)`.

Answer

A, B, C

Explanation

`R = 10m`
`H = 25m`
`(u^(2)sin^(2)theta)/(2g) = 25 rArr usintheta = 10sqrt(5)`
`(dy)/(dx)` at `(0,0)`
`tentheta = 10`
`sintheta = (10)/(sqrt(101))`