1. The problem statement, all variables and given/known data Standing on the ground, you throw a baseball straight upward, releasing the ball at a height of 2.00 meters. The ball travels straight up and then falls(ignoring air resistance). If the ball was in the air for a total of 3.92s, find the maximum height the ball reaches and the speed at which it was thrown. 2. Relevant equations Vf=Vi+at (Vf^2)-(Vi^2)=2ax x=(Vi*t)+(1/2)(a)(t^2) 3. The attempt at a solution I divided my diagram into 3 velocities: velocity initial (Vo):when the ball was released at 2.00 meters velocity 2 (V2): velocity at the top of "bell" which is zero and the velocity once it hit the ground (Vf) Vi=? V2=0 m/s Vf=? i assume Vf is not important I am treating time as a net total (3.92) and therefore tnet=3.92=t1+t2 t1: being the time it take for the ball to reach v2 t2: being the time it takes for the ball to go from v2(top of the bell with 0m/s) to the ground *I treated x the same. Xnet=?=x1+x2 xi:the distance the projectile traveled until velocity became zero. I am assuming x1=2+? x2: the distance the projectile traveled before reaching x=0 meters I played around so much with the equations (substituting equations into each other) that I kept getting 2*G as the acceleration. I know thats not right because the ball was thrown upwards and therefore the velocity must be positive. The answer ended up being: the ball reached 19.8m high and 18.7 m/s as the initial velocity. I've never had a problem like this that hasn't included an initial velocity.