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Vertical motion of a bullet

  1. May 6, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

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

    3. 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 ?)

  2. jcsd
  3. May 6, 2015 #2


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

    That's four significant figures in the result. Again, I count two inputs that are not precisely known.

    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?
  4. May 6, 2015 #3
    In part three, the measurement for the gravitational constant is 9.8 using two significant digits. See what you get when you incorporate that.
  5. May 6, 2015 #4
    I might be incorrect but answer for part 3 should be v2 = 76.0 m/s^2 since it is using two significant digits :)
  6. May 6, 2015 #5


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    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.
  7. May 6, 2015 #6
    I will spend about an hour studying about significant digits, and will come back in hour and reply to you.
  8. May 6, 2015 #7


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    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.
  9. May 6, 2015 #8
    Thank you for clear explanation, I got an idea how all these digits should working. Thank you
  10. May 6, 2015 #9


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    What are the units of velocity? Units are just as important as quibbling over significant figures.
  11. May 6, 2015 #10
    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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