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a certain airplane has a speed of 272.9 km/h and is diving at an angle of θ = 29.0° 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 = 798 m. (a) how long is the decoy in the air? (b) how high was the release point?
(a) number
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(b) number
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Explanation:

Step1: Convert speed to m/s

First, convert the airplane's speed from km/h to m/s:
$272.9\ \text{km/h} = 272.9 \times \frac{1000\ \text{m}}{3600\ \text{s}} \approx 75.81\ \text{m/s}$

Step2: Find horizontal velocity component

Calculate the horizontal component of the velocity:
$v_{x} = v \cos\theta = 75.81 \times \cos(29.0^\circ) \approx 75.81 \times 0.8746 \approx 66.30\ \text{m/s}$

Step3: Calculate time in air (part a)

Use horizontal distance to find time in air, since horizontal motion has constant velocity:
$t = \frac{d}{v_{x}} = \frac{798\ \text{m}}{66.30\ \text{m/s}} \approx 12.04\ \text{s}$

Step4: Find vertical velocity component

Calculate the vertical component of the initial velocity (downward, so positive):
$v_{y} = v \sin\theta = 75.81 \times \sin(29.0^\circ) \approx 75.81 \times 0.4848 \approx 36.75\ \text{m/s}$

Step5: Calculate release height (part b)

Use kinematic equation for vertical displacement, with $a = g = 9.81\ \text{m/s}^2$:
$h = v_{y}t + \frac{1}{2}gt^2$
$h = (36.75 \times 12.04) + \frac{1}{2} \times 9.81 \times (12.04)^2$
$h \approx 442.47 + 710.27 \approx 1152.74\ \text{m}$

Answer:

(a) 12.0 s (rounded to 3 significant figures)
(b) 1150 m (rounded to 3 significant figures)