The basketball passed through the hoop even though it barely cleared the hands of the player b who attempted to block it. suppose that s = 6.5 ft . neglect the size of the ball.

final position based on intial conditions. xf = xo + vxo * t yf = yo + vyo * t - (1/2) * g * t^2 xo = 0.0 ft , yo = 7.0 ft xB = 25 ft , yB = ??? (height B jumps) xH = 30 ft , yH = 10 ft (hoop) initial velocity Vxo = Vcos30; Vyo = Vsin30 gravity: g = 32.2 ft down (thus minus sign in y equation To SCORE: xf = 30 ft; yf = 10 ft solve xf and yf equations for V and t. ... this gets time in air to go 30 ft. For Blocker time to Blocker is: (25/30) tH (tH = time to hoop) substitute this time into yf equation to get yf of person B. alternate: you should have a trajectory equation in your textbook. y as a function of x, Vo and initial angle. no time is needed (basically doing what i have outlined above).

By dividing you can find the answer