Answer: 815.51 m
Explanation:
This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:
[tex]V=V_{o}+gt[/tex] (1)
[tex]V^{2}=V_{o}^{2}+2gy[/tex] (2)
Where:
[tex]V[/tex] is the final velocity of the bullet
[tex]V_{o}=152 m/s[/tex] is the initial velocity of the bullet
[tex]g=-9.8 m/s^{2}[/tex] is the acceleration due gravity, always directed downwards
[tex]t=6.9 s[/tex] is the time
[tex]y[/tex] is the vertical position of the bullet at [tex]t=6.9 s[/tex]
Let's begin by finding [tex]V[/tex] from (1):
[tex]V=152 m/s-9.8 m/s^{2}(6.9 s)[/tex] (3)
[tex]V=84.38 m/s[/tex] (4)
Now we have to substitute (4) in (2):
[tex](84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y[/tex] (5)
Isolating [tex]y[/tex]:
[tex]y=815.511 m[/tex] This is the displacement of the bullet after 6.9 s
