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

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SUMMARY

The maximum altitude of a bullet fired straight up can be calculated using the kinematic equation vf² = vi² + 2ax, where vf is the final velocity (0 m/s at the peak), vi is the initial velocity (3.00 x 10² m/s), a is the acceleration due to gravity (-9.81 m/s²), and x is the displacement (maximum height). By rearranging the equation to solve for x, the maximum height can be determined. The key takeaway is that the displacement represents the bullet's maximum altitude.

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Arbitrary
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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.
 
Last edited:
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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 acceleration being gravity.

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

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