Calculating Maximum Height of a Rocket in Free Fall

AI Thread Summary
The discussion focuses on calculating the maximum height of a rocket that accelerates upward with a constant acceleration of 53.9 m/s² for 5 seconds before entering free fall. The initial calculations yield a height of 673.75 meters and a final velocity of 269.5 m/s at engine cutoff. To find the additional height gained during free fall, the time until the rocket reaches maximum height is determined by dividing the final velocity by the acceleration due to gravity (9.8 m/s²). This time is then used in the distance formula to calculate the height gained against gravity. The total maximum height is the sum of the height during powered ascent and the height gained during free fall.
Bottomsouth
Messages
27
Reaction score
0

Homework Statement



A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 53.9 . The acceleration period lasts for time 5.00 until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.8 .


Homework Equations



Distance = 1/2 at^2
Velocity = u + at


The Attempt at a Solution



So far what I get
distance = 1/2 x 53.9 x 5^2 = 673.75 m
V = 0 + 53.9 x 5 = 269.5 m/s

and from there I seem to be stuck. I don't know what else to do, if there is someone to help guide me or of this ditch I'd appreciate it.
 
Physics news on Phys.org
That's a good start.

Now how much longer until it gets to max height?

Use the same equation that gives you your speed at main engine shut-off. Only this time it's gravity and not rocket power.

Armed with that time figure then using your distance equation how much further it went. Add the 2 heights together. Done.
 
so I multiplied 53.9 by 9.8 and got 528.22, then added that with 676.75 to get 1201.97. Thats not the answer, I am still off somewhere or mis read?
 
Once its done accelerating.

Solve for t
Vf=V0+at
--------------
Vf=0m/s
V0=269.5m/s
a=-9.8m/s
--------------

Distance formula
d=d0+vo*t+1/2at2
--------------
d0 = 673.75m
v0= 269.5m/s
a=-9.8m/s2
--------------
 
Bottomsouth said:
so I multiplied 53.9 by 9.8 and got 528.22, then added that with 676.75 to get 1201.97. Thats not the answer, I am still off somewhere or mis read?

No. You don't multiply the accelerations together. You use the acceleration of gravity this time instead of the acceleration of the rocket over the 5 seconds.

You figured velocity is 269.5. Divide that by 9.8. That's how many seconds it will continue to rise until velocity is back to 0 - time to max height.

Use that time with y = 1/2*g*t2
to find how much farther it goes against gravity.

That height is what you add to your previous height at main engine cut off to get total height.
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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...
Back
Top