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Please check this calculation

  1. Jan 25, 2004 #1
    Can anyone confirm/deny that if an object is dropped (init. velocity=0) that it will take about 2.26 seconds to reach 90 km/h?

    Last edited: Jan 25, 2004
  2. jcsd
  3. Jan 25, 2004 #2
    Assuming no air resistance and an acceleration of 9.81 m/s^2, I get 2.55 seconds.

    Or if the object falls for 2.26 seconds, it will have a velocity of 80 Km/hr.
    Last edited: Jan 25, 2004
  4. Jan 25, 2004 #3
    Can you go through it for me? It's really confusing me as to why I keep getting what I do.
  5. Jan 25, 2004 #4
    Time = Velocity/acceleration

    The units I used were Seconds, m/s, m/s^2

    I converted your value of 90 km/hr to m/s. Divide by 3.6 to convert km/hr to m/s. 90 km/hr = 25 m/s

    I simply plugged in the values into the above formula.

    T = 25 /9.81 = 2.55 seconds.

    Your answer wans't really that far off. What values did you use?
  6. Jan 25, 2004 #5
    Just to start off, we have to assume gravity to be 9.8 in our physics class, but I don't think that can attribute to the difference. Anyhoo, what is wrong with using:

    [tex]V_f = V_i + \frac{1}{2} a t^2[/tex]
    If [tex]V_f[/tex]=25 and [tex]\frac{1}{2}a[/tex]=4.9 and [tex]V_i[/tex]=0, then what should I come out with, or is this the wrong equation?
    Last edited: Jan 25, 2004
  7. Jan 25, 2004 #6
    It's the wrong equation. [itex] 1/2 a t^2[/itex] is an expression of distance traveled under a constant acceleration for a given length of time.

    Since [itex]V_i=0[/itex], your equation simplifies to [itex]V_f = 1/2 a t^2[/itex] which is not correct. [itex]D = 1/2 a t^2[/itex] when intial distance and velocity equal 0. Otherwise,

    [tex]D = D_i+V_i t + \frac{a t^2}{2}[/tex]

    the units on one side of the equation have to match the units on the other side. That's a good way to check for errors in an equation.

    [tex] V_f = V_i + a t [/tex]
    Last edited: Jan 25, 2004
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