# Falling objects problem

## Homework Statement

The problem is: A bullet shot vertically upward leaves the gun at a velocity of 655 m/s. How far above the muzzle will the bullet be 1.0 s after it is fired?

## The Attempt at a Solution

I know that i have the time (1 s) and the velocity (655 m/s). I just don't know how to find the other parts and then which equation to use to find how far above the muzzle the bullet will be.

gneill
Mentor
Clearly the bullet has an initial velocity and will be subject to gravitational acceleration. What kinematic formulas do you know that involve velocity and acceleration?

Clearly the bullet has an initial velocity and will be subject to gravitational acceleration. What kinematic formulas do you know that involve velocity and acceleration?

v = x/t
avg.v = v + initial v/2
v = initial v + at
x - initial x = initial vt + 1/2at^2
v^2 = initial v^2 + 2a(x-initial x)

Nevermind, gneill's suggestion is better.

v = x/t
avg.v = v + initial v/2
v = initial v + at
x - initial x = initial vt + 1/2at^2
v^2 = initial v^2 + 2a(x-initial x)

You know initial velocity, time, and acceleration (gravity, as gneill pointed out). You can solve for final velocity, which will allow you to then solve for distance.

Does that make sense?

Last edited:
gneill
Mentor
v = x/t
avg.v = v + initial v/2
v = initial v + at
x - initial x = initial vt + 1/2at^2
v^2 = initial v^2 + 2a(x-initial x)