Answer:
A. The bomb will take <em>17.5 seconds </em>to hit the ground
B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it
Explanation:
Maverick and Goose are flying at an initial height of
, and their speed is v=688 m/s
When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement
The equation for the height y with respect to ground in a horizontal movement (no friction) is
[1]
With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released
The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time
The range (horizontal displacement) of the bomb x is
[2]
Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:
Setting y=0 and isolating t we get

Since we have 


Replacing in [2]


A. The bomb will take 17.5 seconds to hit the ground
B. The bomb will land 12040 meters on the ground ahead from where they released it
Answer:
33333.35 kg
Explanation:
I got it right on Acellus, rounded to 33300 sigfigs
Answer:
hey but the person at the top is right
This information describes the storm's velocity.
The distance of the object from the lens is 3.12 cm.
<h3>What is magnification?</h3>
Magnification is the process of of enlarging an apparent size of an object.
<h3>Object distance</h3>
The object distance is calculated using the following lens formula;

where;
- M is the magnification
- V is the image distance
- U is object distance

Thus, the distance of the object from the lens is 3.12 cm.
Learn more about object distance here: brainly.com/question/24894435