Show that for a projectile the angle between the velocity and the x-axis as function of time is given
by ` theta_(t) = tan ^(-1)( ( v_(0 y)-g t)/(v_(o x)))`
(b) Shows that the projection angle `theta_0` for a projectile launched from the origin is given by ` theta_(t) = tan ^(-1)( ( 4 h_m)/R)`
where the symbols have their usual meanings.

Explanation

a) Let the principle be fired at an angle `theta` with x-axis. As `theta` depends on t, `theta` (t), at any instant
`tantheta(t)=v_(y)/v_(x)=(v_(oy)-gt)/(v_(ex))`
b) Since, `h_("max") = (u^(2)sin^(2)theta)/(2g)`
and `R=(u^(2)sin2theta)/(g)`
`rArr (h_("max"))/(R) = (u^(2)sin^(2)theta//2g)/(u^(2)sin2theta//g) = (tantheta)/4` (As `sin2theta=2sinthetacostheta)`
`rArr (tantheta)/4= h_("max")/R`
or `tantheta=(4h_("max")/R`
or `theta =tan^(-1)((4h_("max"))/R)`