Maximum height using Newton's second law

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Homework Help Overview

The problem involves determining the maximum height and the time to reach that height for an object thrown vertically upwards using Newton's second law. The object has a specified mass and initial velocity, and the effects of air resistance are to be ignored.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Newton's second law and the role of mass in the equations. Some express uncertainty about how to incorporate mass into their calculations, while others suggest alternative methods such as conservation of energy.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and the requirements to use Newton's second law specifically. Some guidance has been offered regarding the derivation of equations of motion from Newton's second law, but no consensus has been reached on the approach to take.

Contextual Notes

Participants note that the exercise explicitly requires the use of Newton's second law, which is causing some confusion regarding the integration of mass into their calculations.

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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).
 
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Trobocop said:
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?
 
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.
 
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.
 
sjb-2812 said:
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
 
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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