Answer:
10.03 s is the time the decoy is in the air
Explanation:
From the information given we know:
We will need to work in meters and m/s, so [tex]290 \:\frac{km}{h} \cdot \frac{1000 \:m}{km} \cdot \frac{1\:h}{3600\:s} = 80.56 \:\frac{m}{s}[/tex]
The horizontal motion of a projectile is given by:
[tex]x-x_{0}=v_{0x}\cdot t[/tex]
because [tex]v_{0x}=v_0cos(\theta_{0})[/tex], this becomes
[tex]x-x_{0}=(v_0cos(\theta_{0}))\cdot t[/tex]
We have that [tex]\Delta x= 700\:m[/tex], solving for t, we have
[tex]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[/tex]
Answer:
We adopt the positive direction choices used in the textbook so that equations such as
Eq. Â 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. Â to solve for the time:
Δx=(v0​cosθ0​)t⇒t=(80.6m/s)cos(−30.00)700m​=10.0s.
Explanation:
