A certain airplane has a speed of 313.1 km/h and is diving at anangle of = 25.0°

Question

A certain airplane has a speed of 313.1 km/h and is diving at anangle of = 25.0° below the horizontal when the pilotreleases a radar decoy (Fig. 4-37). The horizontal distance betweenthe release point and the point where the decoy strikes the groundis d = 705 m. (a) How long is the decoy in theair? (b) How high was the release point? I cannotfigure out what I'm doing wrong. Its almost the exact same problemas in the textbook I'm just plugging the numbers in. I'm using theformulas. x-x(initial)=(Vinitial cos )t and y-yinitial=(Vinitialsin )(t)-.5gt^2 I cannotfigure out what I'm doing wrong. Its almost the exact same problemas in the textbook I'm just plugging the numbers in. I'm using theformulas. x-x(initial)=(Vinitial cos )t and y-yinitial=(Vinitialsin )(t)-.5gt^2

Explanation / Answer

(a)The time for which the decoy is in the air is
t = (u * sin/g) where u = 313.1 km/h = 313.1 * (1000/3600) m/s = 86.9m/s = 25.0o and g = 9.8 m/s2 (b)The height of the release point is h = (u2 * sin2/2g) (b)The height of the release point is h = (u2 * sin2/2g)

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