Solving Kinematics Problems: Maximum Height, Time, and Velocity Calculations

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Homework Help Overview

The discussion revolves around a kinematics problem involving a bullet fired vertically into the air, with specific questions about maximum height, time to stop rising, and velocity after a certain duration. Participants are analyzing the calculations related to these aspects of motion under gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are discussing the calculations for maximum height using kinematic equations, questioning the initial and final velocities, and the time taken for the bullet to reach its peak. There are also discussions about the accuracy of time used in the velocity calculation after a specified duration.

Discussion Status

Some participants have confirmed the correctness of the original poster's calculations while suggesting adjustments to the initial and final velocities for clarity. There is an ongoing exploration of the implications of using different time values in the calculations, particularly regarding the velocity after 60 seconds.

Contextual Notes

Participants are noting the importance of sign conventions in kinematic equations and the potential impact of using different time values on the results. There is also a mention of the accuracy of measurements and calculations in the context of the problem.

BigCountry
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Good Morning,

Just checking my work, your help would be much appreciated.

A bullet is fired vertically into the air at a speed of 512 m/s.

a) To what maximum height will the bullet does the bullet go?

b) How much time passes before the bullet stops rising?

c) What is the velocity of the bullet after 60 seconds?

My Answers:

a)

Vi = 0, Vf = 512, g = -9.8, find d

Vf^2 = Vi^2 + 2ad
512^2 = 2(-9.8)d
262144 = (-19.6)d
262144/(-19.6) = d
13375 m = d

:smile:


b)

Vi = 0, Vf = 512, g = -9.8, find t

Vf = Vi + at
512 = 0 + (-9.8)t
512/(-9.8) = t
52.24 Secs = t

c)

Vi = 0, g = -9.8, t = 8, find Vf

Vf = Vi + at
Vf = 9.8 * 8
Vf = 78.4 or 78 m/s^2

Thank You
 
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All correct. :smile:
Better not use t=8s in c), but the more accurate 7.76s. Makes a difference of 2m, but who's counting.
 
BigCountry said:
My Answers:

a)

Vi = 0, Vf = 512, g = -9.8, find d

Vf^2 = Vi^2 + 2ad
512^2 = 2(-9.8)d
262144 = (-19.6)d
262144/(-19.6) = d
13375 m = d
I'd say that Vi = 512 m/s, Vf = 0. (This way you won't have to ignore any strange minus signs. :wink: ) Otherwise, looks good.

b)

Vi = 0, Vf = 512, g = -9.8, find t

Vf = Vi + at
512 = 0 + (-9.8)t
512/(-9.8) = t
52.24 Secs = t
Again: Vi = 512 m/s, Vf = 0. Otherwise, looks good.

c)

Vi = 0, g = -9.8, t = 8, find Vf

Vf = Vi + at
Vf = 9.8 * 8
Vf = 78.4 or 78 m/s^2
Huh? t = 60 s, not 8. (Again: Vi = 512 m/s) Note: Signs count!

(Note: Now I see what you did for question c: You started from the top. Good! But take Galileo's advice about accuracy. Note that the velocity is negative.)
 
Last edited:
Thanks Galileo!

Much appreciated!
 

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