# Rocket Height from Launch

• Bottomsouth
In summary, a rocket initially at rest on the ground accelerates upward with constant acceleration of 53.9 m/s^2 for 5 seconds until its fuel is exhausted. It then falls freely, and the task is to find the maximum height it reaches, assuming constant acceleration due to gravity at 9.8 m/s^2 and ignoring air resistance. Using the distance and velocity equations, the height and velocity at main engine shut-off are calculated to be 673.75 m and 269.5 m/s, respectively. To find the maximum height, the time it takes for the rocket to reach 0 m/s is calculated using the velocity equation with acceleration due to gravity, and this time is then used to find the additional

## 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.

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.