Rocket Problem: Kinematics Answer Doesn't Make Sense?

[Sentience]

Homework Statement

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2. The acceleration period lasts for time 10.0 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 \rm{m/s^2} .

Homework Equations

V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

The Attempt at a Solution

I split the problem in 2, with part A being when the rocket is accelerating with fuel and part B being only under the influence of gravity.

During part A, it reaches a height of 1720 M. When I solve for the height in part B using the other data from part A, I get a total distance of 7720 meters, which implies it gained an addiotional 6000 meters.

How can the rocket go further while only under the influence of gravity after the fuel is gone? Is it because its velocity diminishes very gradually?

Perhaps this is just a classical case of science being counter-intuitive, but I was baffled when I calculated that second height.

 

[tedbradly]

Homework Statement

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2. The acceleration period lasts for time 10.0 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 \rm{m/s^2} .

Homework Equations

V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

The Attempt at a Solution

I split the problem in 2, with part A being when the rocket is accelerating with fuel and part B being only under the influence of gravity.

During part A, it reaches a height of 1720 M. When I solve for the height in part B using the other data from part A, I get a total distance of 7720 meters, which implies it gained an addiotional 6000 meters.

How can the rocket go further while only under the influence of gravity after the fuel is gone? Is it because its velocity diminishes very gradually?

Perhaps this is just a classical case of science being counter-intuitive, but I was baffled when I calculated that second height.

Homework Statement

Homework Equations

The Attempt at a Solution

When you take your foot off the gas, how can your car go any farther?

 

[Sentience]

I would expect the car to go further, just not triple the distance it went while I had my foot on the gas.

 

[tedbradly]

I would expect the car to go further, just not triple the distance it went while I had my foot on the gas.

The difference is your car may be going 50mph or 10 mph. The rocket is going 767 mph.

 

[cjl]

Yep. It may be counterintuitive, but that's actually true for pretty much all vertically launched rockets, including the little estes rockets that you can launch from a baseball field. Here's some data from onboard a rocket I launched last year, as well as a picture of the rocket (mainly because I like the picture):

The barometric altitude trace (bluish green) is a bit a bit messy around where the motor burned out due to the supersonic velocity, but you can get a decent idea from this. The red trace shows acceleration, which you can see goes negative around 3.5 seconds, showing the end of the motor burn. At that point, the altitude is somewhere below 3000 feet. Over the following 30 seconds or so, the rocket coasted all the way up to 18,000 feet. It coasted over 5 times farther than it traveled under burn.

As for the reason for this? Anytime a rocket accelerates much faster than it decelerates, this will be the case. Since the change in velocity is the area under the acceleration curve, and the distance traveled is the area under the velocity curve, you can somewhat visualize this. Assuming constant acceleration and no air resistance, if a rocket accelerates from zero to some velocity, and then decelerates back to zero under the influence of gravity, then a graph of velocity vs time will simply be a pair of triangles. The area under the first triangle is altitude gained during burn, and the area under the second will be altitude gained during coast. If the burn happens at 3 times the acceleration of gravity, it will take 1/3 the time of the coast, but the height of the triangle will be the same. This means that it will gain three times as much altitude during coast as during burn in this case. This also scales with rocket acceleration - if the rocket were to accelerate at 10 times gravity during the burn, it would gain ten times the altitude during coast that it did during burn. Air resistance complicates this somewhat (as can be seen in the data above - even though it accelerated at 20 times gravity, it certainly didn't coast 20 times as far as it burned for). However, the same basic reasoning applies.

 

[Sentience]

Thanks for the responses guys. It is starting to make sense now.

Cool stuff cjl! That looks like a lot of fun

 

[Ray Vickson]

Homework Statement

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 34.3 m/s^2. The acceleration period lasts for time 10.0 s until the fuel is exhausted. After that, the rocket is in free fall.

Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 \rm{m/s^2} .

Homework Equations

V = Vo + at
X - Xo = Vot + .5at2
v2 = vo2 + 2a(X - Xo)
X - Xo = .5(Vo + V)t

The Attempt at a Solution

I split the problem in 2, with part A being when the rocket is accelerating with fuel and part B being only under the influence of gravity.

During part A, it reaches a height of 1720 M. When I solve for the height in part B using the other data from part A, I get a total distance of 7720 meters, which implies it gained an addiotional 6000 meters.

How can the rocket go further while only under the influence of gravity after the fuel is gone? Is it because its velocity diminishes very gradually?

Perhaps this is just a classical case of science being counter-intuitive, but I was baffled when I calculated that second height.

When you throw a ball upward, does it rise in the air even though it is being pulled down by gravity and there is no other force acting on it?

PS: do not post notation like at2. If you mean "a times t-squared" you should write it in standard ASCII as at^2, or better still, as a*t^2. Also: v2 --> v^2 and v02 --> v0^2, etc.

RGV