1. The problem statement, all variables and given/known data In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the spike is difficult.Suppose a ball is spiked from a height of 2.30 m with an initial speed of 20.0 m/s at a downward angle of 18.00o. How much farther on the opposite floor would it have landed if the downward angle were , instead, 8.00o? (Source - Fundamentals of PHYSICS, Halliday, Resnick,Walker, 8th edition, chapter 4, QS.33) 2. Relevant equations This is a case of projectile motion , thus, say, initial velocity = u, displacement in vertical direction = h displacement in horizontal direction = s time of flight = t h = uyt + 0.5ayt2 s = uxt + 0.5axt2 3. The attempt at a solution CASE 1:angle = 18o uy = usin(18o) = 6.18m/s -2.3 = 6.18t -4.9t2 t = 1.562s ux= ucos(18o) = 19.02 m/s so, s1 = 19.02 * 1.562 = 29.71 m CASE 2:angle = 8o uy = usin(8o) = 2.78m/s -2.3 = 2.78t-4.9t2 t = 1.025s ux = ucos(8o) = 19.81m/s so, s2= 19.81*1.025 = 20.31m Well, I am confused. Thank you.