Question

A certain airplane has a speed of 272.9 km/h and is diving at an angle of θ = 33.0° 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 = 738 m. (a) How long is the decoy in the air? (b) How high was the release point?

Answer 1

Concept - use equation of kinematics in horizontal reaction to find the time and in vertical direction to find the height at which the decoy was released

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Answer 2

Let's solve the problem step by step:

(a) To find the time the decoy is in the air, we can use the horizontal motion of the decoy. The horizontal distance (d) traveled by the decoy can be calculated using the formula:

d = horizontal speed × time

Rearranging the formula to solve for time:

time = d / horizontal speed

Given values: Horizontal speed = 272.9 km/h (convert to m/s: 272.9 km/h × 1000 m/km ÷ 3600 s/h) d = 738 m

Substitute the values and calculate the time:

time = 738 m / (272.9 km/h × 1000 m/km ÷ 3600 s/h) ≈ 9.13 seconds

So, the decoy is in the air for approximately 9.13 seconds.

(b) To find the height of the release point, we can use the vertical motion of the decoy. The decoy is launched with an initial vertical velocity (Vy) and is subject to free fall due to gravity.

The vertical distance (height) traveled by the decoy can be calculated using the formula:

height = Vy × time - (1/2) × g × time^2

where Vy is the initial vertical velocity, time is the time of flight calculated in part (a), and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Since the initial vertical velocity (Vy) is not given directly, we can find it using trigonometry. The vertical component of the initial velocity can be calculated as:

Vy = horizontal speed × tan(θ)

Given values: Horizontal speed = 272.9 km/h (convert to m/s: 272.9 km/h × 1000 m/km ÷ 3600 s/h) θ = 33.0°

Now, calculate Vy:

Vy = 272.9 m/s × tan(33.0°) ≈ 161.85 m/s

Now, we can calculate the height using the previously calculated time:

height = 161.85 m/s × 9.13 s - (1/2) × 9.8 m/s^2 × (9.13 s)^2 ≈ 720.43 meters

So, the release point was approximately 720.43 meters above the ground.