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Distance travelled in n time when a body is thrown upwards

  1. Apr 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Here is a question in my text book. I saw the answer, but no explanation was there.

    A stone is thrown vertically upward with an initial velocity [itex]v_0[/itex]. The distance travelled by it in time [itex]\frac{1.5v_0}{g}[/itex] is ______________.
    Answer is [itex]\frac{5v_0^2}{8g}[/itex].


    2. Relevant equations
    Max. height reached by a body thrown upwards with an initial velocity v is [itex]\frac{v^2}{2g}[/itex]
    Time taken to reach the max.height with initial velocity v=[itex]\frac{v}{g}[/itex]


    3. The attempt at a solution
    https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-prn2/t1.0-9/1382433_1591373084420691_3016752318298202408_n.jpg

    Time taken to travel AB+BC=[itex]\frac{1.5v_0}{g}[/itex]

    We know that time taken to travel AB=[itex]\frac{v_0}{g}[/itex]

    ∴Time taken to travel BC=[itex]\frac{1.5v_0}{g} - \frac{v_0}{g}[/itex]
    ===================[itex]\frac{v_0}{2g}[/itex]


    Initial velocity at B in BC=u=0

    Acceleration=a=g

    We know [itex]v=u+at[/itex]

    [itex]v=0+g.\frac{v_0}{2g}=\frac{v_0}{2}[/itex]

    [itex]v^2=u^2+2as[/itex]

    [itex]\frac{v_0^2}{2^2}=0^2+2.g.s[/itex]

    [itex]\frac{v_0^2}{4}=2gs[/itex]

    [itex]s=\frac{v_0^2}{8g}[/itex]

    Distance of AB+BC=[itex]\frac{v_0^2}{2g}+\frac{v_0^2}{8g}[/itex]

    ==============[itex]\frac{4v_0^2}{8g}[/itex]
     
  2. jcsd
  3. Apr 28, 2014 #2
    man you made a slight arithmetic mistake in the last line.
     
  4. Apr 28, 2014 #3
    Oh yes! Thanks. I am happy at least I made only arithmetic mistake and not a concept mistake.
     
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