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
[tex]v_x = 500 m/s[/tex]
And the horizontal distance covered by the bullet is
[tex]d=200 m[/tex]
So, since the horizontal motion is uniform, we can immediately find the time it takes for the bullet to cover this distance:
[tex]t=\frac{d}{v_x}=\frac{200}{500}=0.4 s[/tex]
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
[tex]s=ut+\frac{1}{2}at^2[/tex]
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
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity (taking downward as positive direction)
Solving for s, we find
[tex]s=0+\frac{1}{2}(9.8)(0.4)^2=0.78 m[/tex]
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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