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Maximum height using newton's second law

  1. Dec 11, 2011 #1
    a small object with mass m=10gr is thrown vertically upwards from the surface of earth with a velocity v=10m/s

    i need to find the max height(Hmax) (when v=0) and the time(tmax) when the object reaches the max height by using newton's second law only.

    (ignore the effect of air and air resistance to the object).
  2. jcsd
  3. Dec 11, 2011 #2
    What equations do you know that may help here?
  4. Dec 11, 2011 #3
    I'm trying to solve this with [itex]\vec{F}[/itex]=m*a→[itex]\vec{F}[/itex]=m*d2[itex]\vec{r}[/itex]/dt2
    and FB(the weight)=m*g

    so F=-FB*[itex]\hat{k}[/itex]

    but then the mass disappears so i am sure i'm doing this wrong, because the exercise tells me the mass of the object. i have to use it somehow
    if i ignore this, i integrate the above equation then i put u=0 and i find tmax. then by using the equation of motion i find the max height. the numbers are correct because i tried it with energy equations but i am sure it's the wrong way.
  5. Dec 11, 2011 #4
    You don't have to use all the data points all the time... using conservation of energy seems like a good plan for max height to me.
  6. Dec 11, 2011 #5
    yes i know but the exercise says it clearly to use only newton's second law. that's why i'm stuck
  7. Dec 11, 2011 #6
    OK then. What is Newton's second law? F = ma. From this you can derive the equations of motion for constant acceleration. Does that help any?
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