Question

A certain airplane has a speed of 290.0 km/h and is diving at an angle of 0 : 30.0" below the horizontal when the pilot releases a radar decoy (Fig. 4-37). The horizontal distance between the release point and the point where the decoy strikes the ground rs d - 700 m. (a) How long is the decoy in the atr?

Answer 1

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.

Step-by-step explanation:

Answer 2

10.03 s is the time the decoy is in the air

Step-by-step explanation:

From the information given we know:

• 
  `\theta_(0)=-30` because is measured clockwise from horizontal.

• The initial speed of the decoy is the plane's speed at the moment of release
  `v_(0)=290 \:(km)/(h)`

We will need to work in meters and m/s, so
`290 \:(km)/(h) \cdot (1000 \:m)/(km) \cdot (1\:h)/(3600\:s) = 80.56 \:(m)/(s)`

The horizontal motion of a projectile is given by:

`x-x_(0)=v_(0x)\cdot t`

because
`v_(0x)=v_0cos(\theta_(0))`, this becomes

`x-x_(0)=(v_0cos(\theta_(0)))\cdot t`

We have that
`\Delta x= 700\:m`, solving for t, we have

`t=(\Delta x)/(v_(0)cos\theta_(0)) \\t=(700 \:m)/((80.56 \:(m)/(s) )\cdot cos(-30)) \\t= 10.03 \:s`