1. The problem statement, all variables and given/known data A ball is dropped that's 85 feet above the ground. The ball's initial speed is 5 feet/second, and the acceleration of the ball is 32 feet/sec^2. How long does the ball travel before hitting the ground? 2. Relevant equations distance = rate * time 3. The attempt at a solution So the speed of the ball I have as: [t = time] -5 - 32t Which I believe makes sense because the ball is initially traveling @ -5 ft/sec, and after 1 second, the ball is traveling -37 ft/sec, then after 2 seconds the ball is traveling @ -69 ft.sec, etc. I assume we use d = rt (which is distance = rate * time). So the total distance the ball travels is -85 feet which is equal to the total rate of -5 - 32t * time. The equation I set up like this: -85 = (-5 - 32t) * t I set up the quadratic like so: -32t^2 - 5t + 85 = 0 I solved it and got: (1/64) * (sqrt(10905) - 5) which is approx. 1.55 seconds. But the solution says it's 2.15 seconds, and I found out that it's because the solution needs to have taken the average velocity or something, meaning the average of: -5 ft/sec and (-5 ft/sec - (32 ft/sec(t))) So the average turns out to be (-5 ft/sec - 16 ft/sec * t) and I can go on with the rest of the solution and I get about 2.15 seconds as the answer. What I don't understand is why we take the average velocity of the initial and the relative velocities. I thought -5 - 32t was relative to the distance and time that the ball had traveled. For example: At 0 seconds: d = (-5 - 32(0))(0) d = (-5 - 0))(0) d = (-5 ft/sec))(0 seconds) d = 0 ft. So while the ball has not yet traveled, it has a velocity of -5 ft/sec. After 1 second: d = (-5 - 32(1))(1) d = (-5 - 32)(1) d = (-37 ft/sec)(1 second) d = -37 ft. After 1 second, the ball was traveling at -37 ft/sec, and has traveled -37 feet towards its destination so far. After 2 seconds: d = (-5 - 32(2))(2) d = (-5 - 64)(2) d = (-64 ft/sec)(2 seconds) d = -128 ft. The ball is traveling at -64 ft/sec but has already past the -85 ft. mark and presumably through the ground, -43 feet more underground (but that doesn't count right now in this scenario). So when the ball has reached its final destination after 1.55 seconds: d = (-5 - 49.6(1.55))(1.55) d = (-81.88 ft/sec)(1.55 seconds) d = -84.63 ft. The ball was traveling -81.88 ft/sec right when it hit the ground -85 feet down. Just for example's sake, I'll calculate the math for when the ball was halfway to its mark: (-85ft / 2) = (-5 - 32(t))(t) -42.5 ft. = -32t^2 - 5t 0 = -32t^2 - 5t + 42.5 t = 1.07696 seconds So after 1.07696 seconds, the ball was halfway to the ground, and: d = (-5 - 32(1.07696))(1.07696) d = (-39.46272 ft/sec)(1.07696 seconds) d = -42.4997709 feet Which means that halfway to the target (-42.5 feet down), the ball had a velocity of -39.46272 ft/sec at 1.07696 seconds after the ball started to travel. So as a recap: At 0 feet = ball is traveling at -5 ft/sec after 0 seconds of traveling At -37 feet = ball is traveling at -37 ft/sec after 1 second of traveling At -42.5 feet = ball is traveling at -39.5 ft/sec after 1.07696 seconds of traveling At -85 feet = ball is traveling at -81.88 ft/sec after 1.55 seconds of traveling So what is the purpose of taking the average velocity?