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. 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=(v0cosθ0)t⇒t=(80.6m/s)cos(−30.00)700m=10.0s.
Explanation:
