Question

A bullet is fired straight up from a gun with a

muzzle velocity of 152 m/s.
Neglecting air resistance. what will be its
displacement after 6.9s?

Answer 1

Answer: 815.51 m

Step-by-step 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