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Marizza181 [45]
3 years ago
7

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?

Physics
1 answer:
prisoha [69]3 years ago
5 0

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 <em>t, </em>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

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h =ut+  \frac{1}{2} gt^2\\\\h = 0 +  \frac{1}{2} gt^2\\\\h =  \frac{1}{2} gt^2\\\\t= \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 176.4}{9.8} } \\\\t = 6 \ s

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