Question

A certain airplane has a speed of 290.0 km/h and is diving at an angle of 30.0°

below the horizontal when the pilot releases a radar decoy (Fig. 4-33). The

horizontal distance between the release point and the point where the decoy

strikes the ground is d = 700 m. (a) How long is the decoy in the air? (b) How high

was the release point?​

Answer 1

Answer:

Find the volume bounded by xy plane, x²+y²=1 the cylinders and the plane 2x+3y+4z=12

Answer 2

Answer:

We adopt the positive direction choices used in the textbook so that equations such as

Eq.  are directly applicable. The coordinate origin is at ground level directly below

the release point. We write theta0​=–30.0° since the angle shown in the figure is measured

clockwise from horizontal. We note that the initial speed of the decoy is the plane’s speed

at the moment of release: v0​=290km/h, which we convert to SI units=(290)(1000/3600)

= 80.6 m/s.$$

(a) We use Eq.  to solve for the time:

Δx=(v0​cosθ0​)t⇒t=(80.6m/s)cos(−30.00)700m​=10.0s.

(b) And we use Eq. 4-22 to solve for the initial height y0​:

y−y0​=(v0​sinθ0​)t−21​gt2⇒0−y0​=(−40.3m/s)(10.0s)−21​(9.80/s2)(10.0s2)

which yields y0​=897m.

Explanation: