1. The problem statement, all variables and given/known data An object is thrown downward with an initial speed of 4 m/s from a height of 99 m above the ground. At the same instant, a second object is propelled vertically from ground level with a speed of 25 m/s. The acceleration of gravity is 9.8 m/s^2. At what height above the ground will the two objects pass each other? Answer in units of m. 2. Relevant equations None - I believe you make your own equation. Look below to see what I mean. 3. The attempt at a solution Well I tried to find the position equation of both objects, then set them equal to each other to find the time at which their positions are equal. For the top object, since at time 0 his position is 99 m, and his initial velocity is 4 m/s, I found its position equation to be x(t) = -4t + 99. The -4t is negative because its traveling down. For the bottom object, since at the time 0 his position is also 0, it has no constant. His initial velocity is 25 m/s, so I got his position equation to be x(t) = 25t. The 25 t is positive because he is propelled upwards. So I set the two equations equal, and solved for t: -4t + 99 = 25t 99 = 29t t = 3.4138 I plugged this time into one of my position equations to find the position when they meet: x(t) = 25t x(3.4138) = 25(3.4138) x = 85.3448 m This answer seemed to make sense, because since the bottom object is propelled significantly faster than the top object, they should meet somewhere close to the 99 m mark. But low and behold, 85.3448 m is incorrect. Help!!!