When a bady undergoes an oblique projection, maximum range of it is :

Options

A. `(u^(2))/(g)`
B. `(u)/(g)`
C. `(u^(2))/(g^(2))`
D. `(u^(2))/(2g)`

Answer

A

Explanation

`R= (u^(2))/(g) sin 2 theta`
Here `theta = 45^(@)`
`R = (u^(2))/(g) sin 90^(@) = (u^(2))/(g)`