Question

A bullet is fired straight up from a gun with a

Answer 1

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:

$V=V_{o}+gt$ (1)

$V^{2}=V_{o}^{2}+2gy$ (2)

Where:

$V$ is the final velocity of the bullet

$V_{o}=152 m/s$ is the initial velocity of the bullet

$g=-9.8 m/s^{2}$ is the acceleration due gravity, always directed downwards

$t=6.9 s$ is the time

$y$ is the vertical position of the bullet at $t=6.9 s$

Let's begin by finding $V$ from (1):

$V=152 m/s-9.8 m/s^{2}(6.9 s)$ (3)

$V=84.38 m/s$ (4)

Now we have to substitute (4) in (2):

$(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y$ (5)

Isolating $y$:

$y=815.511 m$ This is the displacement  of the bullet after 6.9 s