How High and Fast Does a Bullet Travel When Shot Vertically?

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Homework Help Overview

The discussion revolves around a physics problem involving a bullet shot vertically with an initial speed of 512 m/s. Participants explore the maximum height the bullet reaches, the time it takes to stop rising, and its velocity after 60 seconds, all while considering the implications of significant figures in their calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate maximum height using kinematic equations and question the precision of their significant figures. They also discuss the time taken for the bullet to stop rising and its velocity after a specified time, raising concerns about the accuracy of their inputs and outputs.

Discussion Status

There is an ongoing examination of significant figures and their impact on the results. Some participants provide guidance on how to approach the calculations with respect to significant digits, while others express uncertainty about the correct application of these rules.

Contextual Notes

Participants note that the gravitational constant and the initial velocity are not precisely known, which may affect the accuracy of their calculations. The discussion also highlights the importance of units in the context of the problem.

Glenboro
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Homework Statement


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

1) To what maximum height does the bullet go?

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

3) What is the velocity of the bullet after 60.0 s?[/B]

Homework Equations


1) V2²=V1²+2ad
2) a= (V2-V1)/Δt
3) V2=V1+aΔt

The Attempt at a Solution



1)512^2 = 0^2 + (2)(9.8)(d)
262144 = (19.6)d
262144/19.6 = d
D = 13374. 694m (My teachers says it has too many digit, so should I change to 1.33 X 10^4m ?)

2) T= (0 m/s - 512m/s)/9.8 m/s^2
T = 52.24 s

3) Vf = 512 m/s + (-9.8m/s^2)(60)
Vf = 512 - 588
V2 = 76 m/s^2 downward or opposition direction ( my teacher mentioned about significant digits, so should answer must written as 76.0 m/s^2 ?)

[/B]
 
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Glenboro said:
1)512^2 = 0^2 + (2)(9.8)(d)
262144 = (19.6)d
262144/19.6 = d
D = 13374. 694m (My teachers says it has too many digit, so should I change to 1.33 X 10^4m ?)

What figures in the original equation are not exactly known? I count two.

2) T= (0 m/s - 512m/s)/9.8 m/s^2
T = 52.24 s
That's four significant figures in the result. Again, I count two inputs that are not precisely known.

3) Vf = 512 m/s + (-9.8m/s^2)(60)
Vf = 512 - 588
V2 = 76 m/s^2 downward or opposition direction ( my teacher mentioned about significant digits, so should answer must written as 76.0 m/s^2 ?)
This time the significant figures calculation is trickier. You need to take it step by step. What is the rule for significant figures when adding or subtracting?
 
In part three, the measurement for the gravitational constant is 9.8 using two significant digits. See what you get when you incorporate that.
 
Addem said:
In part three, the measurement for the gravitational constant is 9.8 using two significant digits. See what you get when you incorporate that.
I might be incorrect but answer for part 3 should be v2 = 76.0 m/s^2 since it is using two significant digits :)
 
Glenboro said:
I might be incorrect but answer for part 3 should be v2 = 76.0 m/s^2 since it is using two significant digits :)
How did you conclude that the answer should have two significant digits? It should not. How do you figure that 76.0 m/s2 has two significant digits? It does not.
 
jbriggs444 said:
How did you conclude that the answer should have two significant digits? It should not. How do you figure that 76.0 m/s2 has two significant digits? It does not.
I will spend about an hour studying about significant digits, and will come back in hour and reply to you.
 
If you take ##g = 9.8 ms^{-2}## then that means it could be anywhere between ##9.75 ms^{-2}## and ##9.85 ms^{-2}##. You have the same with ##v## between ##511.5 ms^{-1}## and ##512.5 ms^{-1}##

If you repeat the calculation with those two figures, it will give you an idea of how accurate your answer is. Try maximum ##v## with minimum ##g## and vice versa.

This should show you why you have to cut down the number of digits.
 
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PeroK said:
If you take ##g = 9.8 ms^{-2}## then that means it could be anywhere between ##9.75 ms^{-2}## and ##9.85 ms^{-2}##. You have the same with ##v## between ##511.5 ms^{-1}## and ##512.5 ms^{-1}##

If you repeat the calculation with those two figures, it will give you an idea of how accurate your answer is. Try maximum ##v## with minimum ##g## and vice versa.

This should show you why you have to cut down the number of digits.
Thank you for clear explanation, I got an idea how all these digits should working. Thank you
 
Glenboro said:

Homework Statement


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

1) To what maximum height does the bullet go?

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

3) What is the velocity of the bullet after 60.0 s?[/B]

3) Vf = 512 m/s + (-9.8m/s^2)(60)
Vf = 512 - 588
V2 = 76 m/s^2 downward or opposition direction ( my teacher mentioned about significant digits, so should answer must written as 76.0 m/s^2 ?)

What are the units of velocity? Units are just as important as quibbling over significant figures.
 
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  • #10
SteamKing said:
What are the units of velocity? Units are just as important as quibbling over significant figures.
I usually write down all units, I was just in hurry when I was making a thread. However, I will be more careful to write units everytime
 

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