3. maverick and goose are flying a training mission in their f-14. they are
flying at an altitude of 1500. m and are traveling at 688 m/s (mach 2). they
release their bomb and head for home.

a. how long will it be before the bomb hits the ground

b . how far ( on the ground ) from where they released it will it land ?

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

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

nah.   i'm just sleepy.

momentum of soft ball three times less than baskey ball