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

Shania is trying to determine the oldest geologic unit on this map and the reason to why is she unable to do so is that  the units are all the same age.. when there is the same age, there will be no oldest geologic unit and that i cannot be credible.

The boat's velocity relative to the riverbank can be resolved into two components. the component going north is 2.7 m/s and the component going west is 1.2 m/s.we add these two vectors and then find the magnitude of the resultant to get the speed.the magnitude of the resultant vector from adding can be obtained by the pythagorean theorem.    to two significant figures, the boat's speed relative to the riverbank is 3.0 m/s.