What is the Maximum Altitude of a Bullet Fired Straight Up?

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The discussion centers on calculating the maximum altitude of a bullet fired straight up with an initial velocity of 300 m/s. The user initially struggles with the necessary variables for the equations of motion, particularly the final velocity, time, and displacement. It is clarified that the final velocity at the bullet's peak is zero, which simplifies the problem. The key equation to use is vf^2 = vi^2 + 2ax, where the displacement represents the maximum height. The user acknowledges the oversight and expresses gratitude for the assistance.
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I tried this question multiple times, but now I've become stuck.

A person shoots a gun straight up into the air on New Year's Eve. The velocity at which the bullet leaves is 3.00 x 10^2 m/s. What is the maximum altitude of the bullet?


Here's what I've tried:

I attempted to apply the x=1/2(vi+vf)t formula, but it's missing some variables, namely vf, t, and the displacement. I've also tried to find the final velocity through vf=vi+at, but then I'm missing the time. Then I tried to get the final velocity with vf^2=vi^2+2ax, but I'm now missing the displacement. I must be not seeing something very obvious here.

EDIT: Ahh...gods. I have just realized something. vf = 0 when the bullet reaches the top. Silly me--that means it's not missing any variables.
 
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Arbitrary said:
Then I tried to get the final velocity with vf^2=vi^2+2ax, but I'm now missing the displacement. I must be not seeing something very obvious here.

You're missing the displacement because that is what the problem is asking for. The displacement is the maximum height that the bullet reaches. Use that equation, with accleration being gravity.

EDIT: OK!
 
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Heh, thanks for the help anyways. :)
 
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