1. The problem statement, all variables and given/known data Golf ball struck with a velocity of 66 ft/s. Part a): determine the speed at which the ball hits point B. Part b): determine the time of flight from A to B. 2. Relevant equations For Part b): (Vo)x = Vcos(θ) (Vo)y = Vsin(θ) D = Do + (Vo)x * t Y = Yo + (Vo)y * t - (1/2) * g * t^2 For Part a): Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo) Vx^2 = (Vo)x^2 - 2 * g * (Y - Yo) V^2 = [(V)x^2 + (V)y^2] V = [(Vo)x + (Vo)y] / 2 3. The attempt at a solution If there are more effective and less complicated ways of solving the asked questions please let me know. I solved for time first (part b). -- given an initial velocity Vo in the problem statement of 66 ft/s -- from diagram angle between horizontal and initial direction of the ball 10° + 45° = 55° (Vo)x = Vcos(θ) -------> (Vo)x = 66cos(55) -----> (Vo)x = 37.89 ft/s (Vo)y = Vsin(θ) -------> (Vo)y = 66sin(55) -----> (Vo)y = 54.06 ft/s -- solving for distance d S = So + (Vo)x * t -------> d * cos(10) = 0 + 37.89 * t ----> d = 38.44 * t -- solving for final height of ball (at point B). Y = Yo + (Vo)y * t - (1/2) * g * t^2 -----> d * sin(10) = 0 + 54.06 * t - 16.1 * t^2 -----> 6.68 * t = 54.06 * t - 16.1 * t^2 -- solving for time t t = 2.94s This answer was taken as correct in online program. -- Plugging in t value in d d = 38.44 ft -- Plugging in d value in Y = d * sin(10) Y = 19.62 For my attempt at part a) final speed of the projectile at point B I tried two different formulas, but with no success. Vy^2 = (Vo)y^2 - 2 * g * (Y-Yo) -----> Vy^2 = 54.06^2 - 2 * 32.2 * (19.62 - 0) -----> Vy = 40.73 ft/s Vx^2 = (Vo)x^2 - 2 * g * (S-So) -----> Vx^2 = 37.86^2 - 0 -----> Vx = 37.86 ft/s V^2 = [(V)x^2 + (V)y^2] V = [(Vo)x + (Vo)y] / 2 The fourth equation doesn't make much sense to me, and with the correct values inserted answer resulted incorrect. The third gives me a really high number which also resulted incorrect.