A certain airplane has a speed of 283.6 km/h and is diving at an angle of θ = 29.0° below the horizontal when the pilot releases a radar decoy. The horizontal distance between the release point and the point where the decoy strikes the ground is d = 669 m. (a) How long is the decoy in the air? (b) How high was the release point?

Question

anonymous · Answer

I got 16.2 seconds at a height of 2936.4, but I'm told I'm wrong.

anonymous · Answer

Did you include the y component of the airplane's speed in your calculation?

anonymous · Answer

I think so. To get time, I did: $$669divV _{x}$$

anonymous · Answer

to get height, I did 47.27m/s (sin29)+g(t^2)

anonymous · Answer

Just a minute, I'll draw something up.

anonymous · Answer

ooohhh. didn't divide by two for height.

anonymous · Answer

Ok, I have the initial x velocity at 78.8 m/s. That goes into this: $$s_x=vt$$ or $$78.77\frac{m}{s}t = 669m$$. Now we have the air time, we should be able to get everything else.

anonymous · Answer

$$s_y = v_{yi}t + \frac{1}{2}at^2$$

anonymous · Answer

You shouldn't have to divide by two for the height, because you're only traveling one direction.

anonymous · Answer

8.49 seconds. How did you get your velocity? I just converted km/h to m/s.

anonymous · Answer

I got 744 m for height.

anonymous · Answer

I got 231m, having photo issues with my tablet, just a sec.

anonymous · Answer

$$78.77\left( \sin29 \right)+9.8m/s \left( 8.49^{2} \right)$$

anonymous · Answer

program is saying 231m and 8.41 seconds is wrong

anonymous · Answer

Aw crap. I just realized what I did wrong.

anonymous · Answer

8.49, i mean. Your drawing looks like mine.

anonymous · Answer

|dw:1347939786982:dw| I calculated 29 as the angle from the plane to the ground. It should be on top. The angle to calculate on is 61. That should fix it.

anonymous · Answer

Let me know if that works, gotta go. :(

anonymous · Answer

ahhhh!!!

anonymous · Answer

10 seconds and 897 feet.

anonymous · Answer

Cool. :)