Answer:
The horizontal distance is 478.38 m
Solution:
As per the question:
Initial Speed of the bullet in horizontal direction, [tex]v_{x} = 670 m/s[/tex]
Initial vertical velocity of the bullet, [tex]v_{y} = 0 m/s[/tex]
Vertical distance, y = 0.025 m
Now, for the horizontal distance, 'x':
We first calculate time, t:
[tex]y = v_{y}t - \frac{1}{2}gt^{2}[/tex]
(since, motion is vertically downwards under the action of 'g')
[tex]0.025 = 0 - \frac{1}{2}\times 9.8t^{2}[/tex]
[tex]t = \sqrt{0.05}{9.8} = 0.0714 s[/tex]
Now, the horizontal distance, x:
[tex]x = v_{x}t + \frac{1}{2}a_{x}t^{2}[/tex]
[tex]x = v_{x}t + \frac{1}{2}0.t^{2}[/tex]
(since, the horizontal acceleration will always be 0)
[tex]x = 670\times 0.714 = 478.38 m[/tex]
