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Finding the angle of release from a moving plane that drops an object X distance away 
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#1
Jun1307, 08:18 PM

P: 85

1. The problem statement, all variables and given/known data
A plane flies horizontally @ 299 m/s relative to ground. Altitude: 3450m, level terrain; neglect air resistance. 1: How far from the point vertically under the point of release does the object land (in meters)? 2: At what angle must the bomb be released so that it hits the target at time of release (in °)? 2. Relevant equations t = (sqrt(Voy^2 + 2gh)  Voy) / g Range = Vox(t) 3. The attempt at a solution Using the above, I correctly figured 7933.83 m for the range. It's the angle part that's screwing me. Now, because the ycomponent of that object's velocity is zero at release, I was able to answer #1 without breaking anything down. But I have not a clue as to how to find that damn angle. 


#2
Jun1307, 08:31 PM

Mentor
P: 41,568

I assume they want the bomb to hit the target nose first. What angle does the velocity make when it hits the target? (Find the velocity components.)



#3
Jun1407, 04:19 PM

P: 85




#4
Jun1407, 04:30 PM

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P: 3,017

Finding the angle of release from a moving plane that drops an object X distance away
To me the second question does not make sense. First, the bomb cannot hit the target at the time of release! Second, I have the *feeling* that they mean to ask the angle at which it hits the target, not at which it is released, but I can't tell for sure. Third, if they want an angle of release, they have to give more information (how far the target is when th ebomb is released and more stuff as well). is this the way is exactly phrased?? 


#5
Jun1407, 04:32 PM

P: 85




#6
Jun1407, 04:49 PM

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Thanks
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#7
Jun1407, 04:58 PM

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You just have to find the angle at which the target appears from the bomb's point of view when the bomb is released. Dick explained it. It has nothing to do with physics at this point, it's just trigonometry. 


#8
Jun1407, 05:27 PM

P: 85

tan^1 (7933.8264 / 3450) = 66.498° it is. Thanks. 


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