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

Answer:

10.03 s is the time the decoy is in the air

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 $v_{0}=290 \:\frac{km}{h}$

We will need to work in meters and m/s, so $290 \:\frac{km}{h} \cdot \frac{1000 \:m}{km} \cdot \frac{1\:h}{3600\:s} = 80.56 \:\frac{m}{s}$

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

$t=\frac{\Delta x}{v_{0}cos\theta_{0}} \\t=\frac{700 \:m}{(80.56 \:\frac{m}{s} )\cdot cos(-30)} \\t= 10.03 \:s$