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?
The Attempt at a Solution
Okay, step one is to find the velocity in the x and y directions, here is my attempt:
So we have the following knowns:
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
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!