Question

Show that projection angle `theta_(0)` for a projectile launched from the origin is given by:
`theta_(0)=tan^(-1)[(4H_(m))/(R)]` Where `H_(m)=` Maximum height, R=Range

Answer 1

`H_m = (u^2 sin^(2) theta_0)/( 2g) , R= (u^(2))/( g) sin 2 theta_0 = (2u^(2)) /(g) sin theta _(0).cos theta_0`
`(H_m) /(R ) = (1)/(4) tan theta _(0) rArr theta_0 = tan ^(-1) ((4H_m)/(R ))`