1. The problem statement, all variables and given/known data A ball is projected horizontally from the edge of a table that is 1.02 m high, and it strikes the floor at a point 1.28 m from the base of the table. What is the initial speed of the ball? Correct, computer gets: 2.81e+00 m/s How high is the ball above the floor when its velocity vector makes a 43.6° angle with the horizontal? 2. Relevant equations For the first question, I already figured out that the initial speed of the ball was 2.81 m/s and that the time it took for the ball to reach the ground 1.28 m away from the base of the table to be 0.456 seconds. 3. The attempt at a solution For the second related question, I used the following methods to try and solve: At the top (edge of table) before the ball begins to fall, the horizontal angle to the ball would be as follows: tan(theta) = 1.02/1.28 theta = 38.55 degrees The change in angle degrees from that first point to the point when the ball reaches the ground would be 90-38.55 = 51.45 degrees. Since we want a change of only 43.6-38.55 = 5.05 degrees, I used a simple ratio as follows because acceleration due to gravity is constant (the change in degrees per time interval should stay constant): 51.45 degrees/0.456 seconds = 5.05 degrees/x x = 0.045 s I then plugged in this time interval into the following equation: x(t) = x + vt + 0.5at^2 x(t) = 0 + 0 + 0.5(-9.8 m/s^2)(0.045 s)^2 x(t) = -9.81 x 10^-3 m Finally, to get height: 1.02 m - (9.81 x 10^-3 m) = 1.01 m However, this answer is apparently wrong. Any suggestions?