The trajectory of a projectile is given by `y=x tantheta-(1)/(2)(gx^(2))/(u^(2)cos^(2)theta)`. This equation can be used for calculating various phenomena such as finding the minimum velocity required to make a stone reach a certain point maximum range for a given projection velocity and the angle of projection required for maximum range. The range of a particle thrown from a tower is define as the distance the root of the tower and the point of landing.
In the previous problem, what should be the corresponding projection angle.

Options

A. `tan^(-1) (1//2)`
B. `tn^(-1)(1//3)`
C. `tan^(-1)2`
D. `tan^(-1)3`

Answer

C

Explanation

`H = R tan theta - (1)/(2) g. (R^(2))/(u^(2)) (1 + tan^(2) theta)`
`dy // d theta = 0`
`tan theta = (u^(2))/(Rg) = (100)/(5 xx 10) = 2 rArr theta = tan^(-1) 2`