1. The problem statement, all variables and given/known data A. If a bullet that leaves the muzzle of a gun at 281 m/s is to hit a target 77.0 m away at the level of the muzzle, the gun must be aimed at a point above the target. How far above the target is this point? B. The coach throws a baseball to a player with an initial velocity of 27.0 m/s at an angle of 38.0° with the horizontal. At the moment the ball is thrown, the player is 73.10 m from the coach. At what velocity must the player run to catch the ball at the same height that the ball was released? Your result should be negative if the player is running towards the coach and positive if he is running away. 2. Relevant equations A. d = v/t B. R = (V0^2 * sin(2 * theta)) / g t = 2v0y / g 3. The attempt at a solution A. I'm really not too sure what to do with this one after I find the time. The only thing I could think of what be to relate the acceleration to that time, but I didn't come out with the right answer. B. I'm pretty sure I have the right answer on this one, but the computer thinks otherwise. I found the range of the ball given the above equation, and then found the time after finding the velocity of v0y. From here, I subtracted the 73.10 from my range, and then divided that number by the time I got for the average velocity. Keep in mind these were evidently wrong, but here were my numbers: R = 72.1781 t = 3.3924 Velocity of the player = -0.272m/s Any help would be greatly appreciated.