If `R` is the horizontal range for `theta` inclination and `h` is the maximum height reached by the projectile, Then maximum range is

Options

A. `R^(2)/(h) +2h`
B. `R^(2)/(8h)+2h`
C. `R^(2)/(8h)+8h`
D. `R^(2)/(h)+h`

Answer

B

Explanation

We know that horizontal range, `R=(u^(2)sin2theta)/g`
and maximum height `h=(u^(2)sin^(2)theta)/(2g)`
`:. (R^(2))/(8h)+2h([(u^(2)sin 2 theta)/(g)]^(2))/(8[(u^(2) sin^(2) theta)/(2g)])+2[(u^(2) sin^(2) theta)/(2g)]`
`=(u^(4)(2sin thetacos theta)^(2))/(g^(2)xx8(u^(2)sin^(2)theta)/(2g))+(u^(2)sin^(2)theta)/g`
`=u^(2)/(g)(cos^(2)theta+sin^(2)theta)=u^(2)/g=R_(max)`