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
![v_x = 500 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20500%20m%2Fs)
And the horizontal distance covered by the bullet is
![d=200 m](https://tex.z-dn.net/?f=d%3D200%20m)
So, since the horizontal motion is uniform, we can immediately find the time it takes for the bullet to cover this distance:
![t=\frac{d}{v_x}=\frac{200}{500}=0.4 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv_x%7D%3D%5Cfrac%7B200%7D%7B500%7D%3D0.4%20s)
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
![s=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
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
![s=0+\frac{1}{2}(9.8)(0.4)^2=0.78 m](https://tex.z-dn.net/?f=s%3D0%2B%5Cfrac%7B1%7D%7B2%7D%289.8%29%280.4%29%5E2%3D0.78%20m)
So, the initial height of the bullet was 0.78 m above the ground.
Learn more about projectile motion here:
brainly.com/question/8751410
#LearnwithBrainly