How High Will a Bullet Travel Fired Straight Up at 460 m/s?

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A bullet fired straight up at a velocity of 460 m/s will take time to reach its highest point, which can be calculated using the kinematic equation vf = vi + at, where the acceleration is -9.8 m/s² due to gravity. By solving for time, the bullet's ascent duration can be determined. Once the time to reach the peak is known, the maximum height can be calculated using the equation d = vt + 0.5at². The acceleration due to gravity is essential for these calculations. Ultimately, the bullet's trajectory can be accurately assessed by applying these formulas.
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A bullet is fired straight up with a muzzle velocity of 460 m/s. how long will it take it to reach its highest point and how high will that be? (The air resistance may be neglected)

i've been busting my brain over this question for almost an hour, and i just can't find a formula that will help me find this that only has one of: Distance and Time
 
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I think you should be able to find the time using the kinematic equation:

vf = vi + at

solve it for t, the problem gives you the acceleration, the final velocity and the initial velocity.

then, after you've found the time at the top of its flight, you can use

d = vt + .5at^2

to find the distance traveled.
 
what's the acceleration? it doesn't state it in the equation. or is it -9.8 m/s^2
 
right on dude
 
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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