Note: I figured this out. See my follow-up. Thanks. 1. The problem statement, all variables and given/known data Two tall buildings of equal height stand with vertical walls 40m apart. A stone is thrown from the top of one at an angle of 25 degrees below the horizontal and directly at the other building with a velocity of 18 m/s. How far down from the top does it strike the opposite wall? 2. Relevant equations 3. The attempt at a solution Note, first of all, that the picture it shows in my book has the stone coming off the very top of the building. There's no one throwing the stone, so it's being thrown at y=0, where the very top of the building is 0. I've worked this problem several times, and I keep getting the same (wrong) answer, according to the answers at the end of my book. Here's my process: 25 degrees down, with an initial velocity of 18 m/s means the horizontal component of the velocity is: 18 m/s * cos(-25) = 16.31 m/s The vertical component is 18 m/s * sin(-25) = -7.607 m/s So far so good? The stone will travel 40 m horizontally at 16.31 m/s, which means the time is 40 m / 16.31 m/s = 2.45s So, with that, I can use my initial vertical velocity of -7.607 m/s, acceleration of -9.81 m/s^s, and time = 2.45s, to calculate the y. y = (-7.607m/s)*(2.45s) + (1/2)*(-9.81m/s^2)(2.45^2) = -48.08 m That's not the correct answer, according to my book. Can anyone spot what I'm doing wrong? (The book says the answer is 41.8m, btw) Thanks.