I am looking at a paper which is obtaining initial velocity from displacement to derive an equation for initial velocity based on height of the basketball hoop-height of ball release.

To get the equation displacement of the ball in the vertical motion, it will use the equation (2): y = yo+ Voy 't - 7 gt? h' = 0+ vo sin 0 (t, +t2 ) - 58(t, +tz)? Then, h' is simplified and reduced to (3) as below: h' = 8 ( t , + t 2 ) [2t , - (t, + +2 ) ] (3) Substituting equation (1) and equation (2): X X h' 2t 2 vo COS 0 Vo COS 0 Vo = gx? (tan? 0 + 1 V 2 (x tan 0 - h') (4) To solve the problem, v, needs to be differentiated with respect to 0. Simplify by using trigonometric identities tan' 0+ 1 = sec 0 and chain rule applied: df (u) df du dv. gx sec' 0 = xtan 0 - 2 h' tan 0 - x dx dx dx do 2xtan0 - 2 h