Maximum height using Newton's second law

AI Thread Summary
To find the maximum height (Hmax) and the time (tmax) for a small object with a mass of 10 grams thrown vertically upward at 10 m/s using Newton's second law, the relevant force is the weight (FB = mg). The acceleration due to gravity (g) acts downward, resulting in a net force of F = -mg. By applying F = ma, the mass cancels out, leading to the equations of motion for constant acceleration. The maximum height can be calculated by integrating the equations of motion, while ensuring that the calculations adhere strictly to Newton's second law as specified in the exercise. The discussion emphasizes the importance of following the exercise's constraints and using the correct approach to derive the necessary equations.
Trobocop
Messages
3
Reaction score
0
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).
 
Physics news on Phys.org
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 \vec{F}=m*a→\vec{F}=m*d2\vec{r}/dt2
and FB(the weight)=m*g

so F=-FB*\hat{k}

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?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top