1. The problem statement, all variables and given/known data This is Chapter 4 problem 45 of Halliday/Resnick/Walker Fundamentals of Physics 8th Edition. I can get the right answer (I'll show you...) but when I apply this answer to the problem I get a weird result for another derived value in the problem: the horizontal displacement of the projectile at the time it hits the target. I am probably missing something obvious or maybe subtle. Problem The Archer fish is able to shoot drops of water at insects sitting on branches near the water, and knock them off and eat them. A particular fish is at a stright-line distance of .9m from an insect. The angle from the surface of the water up to the line of sight to the insect is 36 deg. The fish shoots the drop at a velocity (Vo) of 3.56m/s. The task is to find the launch angle the fish should shoot the drop to hit the insect, subject to the constrain that the drop should be at the top of its parabolic path when it hits the insect. 2. Relevant equations These are described below. 3. The attempt at a solution The distance from the fish to the insect is .9m and the angle up to the insect is 36 degrees, so we have a right triangle with a hypotenuse of .9m and one angle of 36 deg. So to get the height of the insect above the water (h) we know sin(36)=h/.9m, so h=.53m. We can also get the horizontal distance (x) from the fish to the point under the insect since cos(36)=x/.9m, so x=.73m. To get the launch angle to shoot the drop (B) we use two equations. The first is the projectile motion equation for the height y: y=Vo sin(B)t-(g/2)t^2, with Vo=initial velocity, B the launch angle, t the time and g the free-fall acceleration (9.8m/s^2). The second is the equation for the y-component of velocity Vy=Vo sin(B)- gt. We know that at the top of its parabolic path Vy=0,and the problem says this is when the drop hits the fish. So by the second equation Vo sin(B)-gt=0. So t=Vo sin(B)/g. We plug this expression for t into the first equation for the height and after simplifying get y= Vo^2 (sin(B))^2/(2g). We know y is .53 by the geometry of the problem. So we can solve this equation for B and get B=arcsin ( (2g y)/Vo^2)^(1/2). When you do this out you get B=64.87 degrees. This is the answer in the book. We can get the time to hit the insect by using t=Vo sin(B)/g, i.e. t=(3.56m/s)(sin(64.87)/(9.8m/s^2)= .33 seconds. We can check this solution by plugging in this t=.33s and B=64.87 deg into y=Vo sin(B)t-(g/2)t^2 and get .53m which is the height of the insect by the geometry of the problem. NOW HERE IS WHERE I GET CONFUSED: If the launch angle is 64.87 deg and the initial velocity is given as 3.56m/s (which we used to correctly solve the problem) then the component of the initial velocity in the x-direction is 3.56 cos (64.87)=1.51m/s. Since this is projectile motion this x-component of the initial velocity (call it Vx) does not change with time. So we should be able to use this Vx and the time we hit the drop (.33s) to get the horizontal distance from the fish to the point underneath the drop. So we just do velocity times time: (1.51m/s)(.33s)=.49m. But the geometry of the problem showed us that the horizontal distance from the fish to the point under the insect is .73m, not .49m. So the projectile motion equations give the correct y but not the correct x for the point where the drop hits the insect. I'm probably missing something obvious...any help appreciated.