A military helicopter dropa BOMB

AI Thread Summary
A military helicopter accidentally drops an unarmed bomb from 250 meters while flying at 65.0 m/s. The bomb takes approximately 7.14 seconds to reach the ground, traveling a horizontal distance of 0.464 km during its fall. The horizontal component of the bomb's velocity just before impact remains 65 m/s, while the vertical component is calculated to be around 70.04 m/s. The discussion emphasizes the independence of horizontal and vertical motions, noting that the helicopter will remain directly above the bomb until it hits the ground. Understanding these principles is crucial for solving similar physics problems effectively.
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A military helicopter on a training mission is flying horizontally at a speed of 65.0 m/s and accidentally drops a bomb (fortunately not armed) at an elevation of 250 m. You can ignore air resistance.

How much time is required for the bomb to reach the earth?(s)

How far does it travel horizontally while falling?(km)

Find the horizontal and vertical components of its velocity just before it strikes the earth.(m/s)

If the velocity of the helicopter remains constant, where is the helicopter when the bomb hits the ground? height above ground(m):confused: :mad:
 
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You need to show us some work first before we can help you. What have you tried so far, where are you getting stuck?
 
Use regular kinematics formulas and you'll be fine
 
so far i have founf the time for the bomb to fall which is 7.14s using x=VT+gT^2/2
I also found how far it travels which is .464km

But i don't know how to find the vertical and horizontal components and where the helicopter is when the bomb hits the ground
 
Now calculate the horizontal velocity from horizontal distance and time

and

the vertical velocity from vertical distance and time
 
would the horizontal componant be 65m/s and the vertical component be 70.04m/s

I am not sure i f it is right
 
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Without grabbing a calculator, 70.04 sounds reasonable after doing some mental math (I usually use g=10m/s^2; after 7 seconds, the velocity is 70 m/s down, average velocity is 35 m/s down and distance is 7s*35m/s = 245 meters down; so your answer sounds quite reasonable.)

You could have also used v_f^2 = v_i^2 + 2ad for the velocity calculation, rather than relying on that intermediate step (although the time was asked for, so you had to find it anyway.)

If you have a physics teacher with a nice sense of humor, point out that the helicopter is going to follow the bomb all the way to the ground if there's no air resistance :)
 
view the video of the B-52 dropping bombs

I'm on dial-up, so I'm not going to wait 14 minutes for this video to load. However, I think it's the correct video that I use in class to demonstrate the answer to your question. Hopefully it helps you understand the independence of horizontal and vertical velocities. i.e. note where the bombs are as they fall in relationship to where they came out of the B-52. This isn't a star trek model where everything is hanging from strings and the background is moving... :)

http://www.danshistory.com/b52.html"
 
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