(a) Show that for a projectile the angle between the velocity and the x-axis as function of time is given
` by ` theta_(t) = tam ^(-1) ( 9 4 h_m)/R)`
where the sybols have thir usual meanings.

Explanation

(a) Let ` v_(ox)` and ` v_(oy)` be the initial component velocity of the projectile at (O) along (OX) direction ans (OY) direction respectively, where (X) is horizontal and (OY) is vetical. Let the projectioe go from (O) from (P) in time (t) and ` _x. v_y` be the component velocity of projectile at (P) along horizontal and vertical directions respectively.
Then, ` v_y = v_(oy) -gt`
and ` v_x = v(ox)`
If ` theta` is the angle wich the resultant velocity` vec v` makes with horizontal direction, then
` tan theta = v_y /u_x = ( v_(oy)- gt)/v_(ox)`
or ` theta =tan^(-1)` ((v_(oy) - gt)/(v_(ox))`
(b) In angular projection,
Maximum vertcal height, ` h_m = ( u^2 sin ^2 theta_0)/( 2 g0`
Horizontal range, ` R= ( u^2 sin 2 theta_0)/g = u^2/g 2 sin theta-0 cos theta-0`
So, ` h_m/R = 9tan theta_0)/4 or tan theta_0 = ( 4h_m)/R`
or ` theta_0 tan^(-1) ( (4 h_m) /R)`.