**1) The bullet remained in the air for 0.4 seconds.**

**2) The bullet was fired from 0.78 m above the ground**

**Explanation:**

1)

The motion of the bullet in the problem is a **projectile motion,** so it follows a parabolic path, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion, with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the horizontal motion of the bullet. In fact, we know that:

- The horizontal velocity of the bullet, which is constant, is

And the horizontal distance covered by the bullet is

So, since the horizontal motion is uniform, we can immediately find the time it takes for the bullet to cover this distance:

So, the bullet remained in the air for 0.4 s.

2)

To solve this part, we analyze the vertical motion of the bullet, using the suvat equation

where

s is the vertical displacement

u is the initial vertical velocity

t is the time of flight

a is the acceleration

For the bullet here we have:

u = 0 since the bullet is fired horizontally

t = 0.4 s is the time of flight

is the acceleration of gravity (taking downward as positive direction)

Solving for s, we find

So, the initial height of the bullet was 0.78 m above the ground.

Learn more about **projectile motion** here:

brainly.com/question/8751410

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