Finding the maximum height of a rocket

AI Thread Summary
The discussion focuses on calculating the maximum height of a rocket that accelerates upward with a constant net acceleration until its fuel is exhausted. The correct formula for maximum height, considering both the upward acceleration and the effect of gravity, is expressed as (at)^2/2g + 1/2at^2. This equation accounts for the initial velocity of the rocket and the gravitational pull acting on it after the fuel runs out. Participants clarify that the rocket's trajectory is parabolic, meaning it continues to rise briefly after fuel depletion before descending. The conversation emphasizes the importance of including both acceleration and gravity in the height calculation.
hieule
Messages
5
Reaction score
0
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t_1, when the fuel is exhausted.

Find the maximum height H that the rocket reaches (neglecting air resistance).
Express the maximum height in terms of a, t_1, and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity.

my answer is:
(at)^2/2g +1/2at^2

is that correct yet?
 
Last edited:
Physics news on Phys.org
hieule said:
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant net acceleration a, until time t_1, when the fuel is exhausted.

Find the maximum height H that the rocket reaches (neglecting air resistance).
Express the maximum height in terms of a, t_1, and/or g. Note that in this problem, g is a positive number equal to the magnitude of the acceleration due to gravity.

my answer is:
(1/2)*(g)*(t1)^2

but that is incorrect

Your equation only shows the distance as far as when the fuel is gone. A rocket's path is a parabola.. when it shoots up and then runs out of fuel, it will go up a little more and then start to fall.
 


Yes, your answer is correct. The maximum height reached by the rocket can be expressed as (at)^2/2g +1/2at^2, where a is the constant acceleration, t is the time at which the fuel is exhausted, and g is the acceleration due to gravity. This formula takes into account the initial velocity of the rocket (which is zero) and the fact that the rocket is accelerating both due to its own engine and the force of gravity.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top