1. The problem statement, all variables and given/known data A certain airplane has a speed of 267.2 km/h and is diving at an angle of θ = 28.0° below the horizontal when the pilot releases a radar decoy (Fig. 4-37). The horizontal distance between the release point and the point where the decoy strikes the ground is d = 777 m. (a) How long is the decoy in the air? (b) How high was the release point? 2. Relevant equations d=Vi*t+1/2a(t^2) 3. The attempt at a solution Okay, step one is to find the velocity in the x and y directions, here is my attempt: Vy=sin(-28)*267.2= -125.4428016 Vx=cos(-28)*267.2= 235.9235968 So we have the following knowns: V=267.2m/s Vy=-125.4428016m/s Vx=235.9235968m/s Dx=777m Ax=0 Ay=-9.8m/s^2 Dy=??? T=??? Okay, cool, plenty of information. Let's solve for T first so we can find Dy afterwards: Dx=VxT + 1/2Ax(T^2) 777=(235.9235968)(T) + (1/2)0(T^2) 777=(235.9235968)(T) + 0 T=777/(235.9235968) T=3.293439107 ----STOP---- Now at this point, I went on to solve for Dy, but we're gonna stop here for the following reason: when I submitted my answer for T to the wileyplus online system, it told me my answer was wrong. What did I do wrong? Everything seems to make sense. Please help!