What is the equation for finding the displacement of a falling object?

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To find the displacement of a bullet shot vertically upward with an initial velocity of 655 m/s after 1 second, the relevant kinematic equations must be applied, considering gravitational acceleration. The key equations discussed include x - initial x = initial vt + 1/2at^2 and v = initial v + at, which relate velocity, time, and displacement. The initial velocity, time, and acceleration due to gravity are known, allowing for the calculation of final velocity and distance. The fourth equation mentioned is particularly useful for solving the displacement. Understanding these kinematic relationships is essential for accurately determining the bullet's position after 1 second.
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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?

Homework Equations





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.
 
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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?
 
gneill said:
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.

lindseyam said:
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:
lindseyam said:
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)

Your fourth equation looks promising...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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