A particle is projected under gravity with velocity `sqrt(2ag)` from a point at a height h above the level plane at an angle `theta` to it. The maximum range R on the grond is :

Options

A. `sqrt((a^(2)+1)//h)`
B. `sqrt((a^(2) h)`
C. `sqrt((ah)`
D. `2sqrt(a(a+ h))`

Answer

d

Explanation

Coordinate of point P are (R,-h)
Hence `-h=R tan theta-(gR^(2))/(2(2ga))(1+tan^(2) theta)`
or `R^(2)tan^(2) theta-4aRtan theta+(R^(2)-4ah)=0`
For `theta` to be real.
`(4aR^(2))ge4R^(2)(R^(2)-4ah)`
or `4a^(2)ge(R^(2)-4ah)`
or `R^(2)le4a(a+h)`
`R_(max)=2sqrt(a(a+h))`