Maximum height using newton's second law

  • Thread starter Trobocop
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  • #1
Trobocop
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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).
 

Answers and Replies

  • #2
sjb-2812
445
5
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).

What equations do you know that may help here?
 
  • #3
Trobocop
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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.
 
  • #4
sjb-2812
445
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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.
 
  • #5
Trobocop
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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.

yes i know but the exercise says it clearly to use only newton's second law. that's why i'm stuck
 
  • #6
sjb-2812
445
5
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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