Acertain airplane has a speed of 303.9 km/h and is diving at an angle of θ = 30.0° below the horizontal when the pilot releases a radar decoy (see the figure). the horizontal distance between the release point and the point where the decoy strikes the ground is d = 610 m. (a) how long is the decoy in the air? (b) how high was the release point?

(a) How long is the decoy in the air:

the distance the decoy traveled is given by the formula
x=v₀x*t
v₀x= velocity of the plane=303.9kmph=84.416667mps
x= The horizontal distance between the release point and the point where the decoy strikes the ground=610m
t=time
plug the values
time equals 7.22sec

(b) How high was the release point:

tanθ= opposite/adjacent 
here the opposite side is the height=h
adjacent is the range of the decoy=610 m

θ= 30°
and tan30° =0.57
plugging the values

height= 347.7m

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