# Experimenting with a Solid-Fueled Rocket: Flight Analysis

• Naeem
This may not be the easiest way to solve the problem, but it can be solved this way. In summary, the maximum altitude the rocket reaches is 48.97 meters. The rocket's acceleration 6.5 seconds after the launch is 7.54 m/s^2. The total time of flight is 10.84 seconds. The rocket's total displacement during the entire flight is zero, as it ends up in the same place it started. The only force acting on the rocket as a free fall particle is gravity, with a magnitude of 9.8 m/s^2. The time it takes to get to the top is 8.15 seconds. The rocket
Naeem
Q. In an experiment in physics class, a model solid-fueled rocket is fired vertically with an acceleration of 3 m/sec2 for 5 seconds. After that time, it's fuel is exhausted and it continues upwards as a free fall particle. (Take upwards to be the positive direction.)

a) What is the maximum altitude the rocket reaches?

Ans. x-x0=vo(t)+(1/2)a(t)^2
where x0=0
vo=0
a=3 m/sec^2
t=5

I get x= 37.5 meters

v=0, a=-9.81m/s/s t=5 sec

Using v = v0 + at, we get

v0 = 49.05 m/s^2

While the rocket engine is firing a= 3. v0= 0 so v= 0+ 3(5)= 15 and you can use THAT as v0 after the engine shuts down.

AFTER the rocket engine has shut off, the only acceleration is that of gravity: -9.81 so v= 15- 9.81t (t is now measured from engine cutoff, not initial launch). From that we get h= 15t- 4.9t2+ 122.5 (since the height was 37.5 m at engine cutoff, t=0).

The maximum height occurs when v= 0: 15- 9.81t= 0 gives t= 15/9.81= 1.52 seconds. That height will be h= 15(1.52)- 4.9(1.52)^2+ 37.5= 48.97 m.

b) What is the rocket's acceleration 6.5 seconds after the launch?

Here is where I am stuck. Any ideas...

c) What is the total time of flight?

Some ideas:

Finally, the flight will be over when the rocket hits the ground:
h=0. Solve 15t- 4.9t2+ 37.5= 0 to find the time AFTER CUTOFF that that occurs. You will need to add the inital 5 seconds to find the length of the entire flight.
I tried to do this but my answer is wrong. ( The computer does not take it )

d) What is the rocket's total displacement during the entire flight?

The answer is zero, but can anybody explain why?

Displacement is not the total distance moved, but a vector of the distance from the starting point to the endpoint. Since it ended up in the same place it started (on the ground), the displacement is 0.

Look up the exact definition of displacement in a physics textbook if that doesn't quite make sense.

Can anybody help me on part b and c... Plz

b) It is no longer accelerating upwards, what forces are acting on it as a free fall particle? Whats the magnitude of this force?

c) Find the time it takes to get to the top. The time for it to fall will just be a freefall problem with the kinetic equation for freefall.

$$t = \sqrt{\frac{2h}{g}}$$

b) Acceleration due to gravity is acting only

I tried doing, v = v0 + at

v0 =0

a = v / t
v = 49.05 m/s
t = 6.5 s

a= 7.54. which is wrong?

c) I used t = sqrt (2h/g) 3.15 seconds, + the Orignal t = 5 seconds = 8.15 seconds which is wrong.

Don't know what is wrong here...

Naeem said:
b) Acceleration due to gravity is acting only

I tried doing, v = v0 + at

v0 =0

a = v / t
v = 49.05 m/s
t = 6.5 s

a= 7.54. which is wrong?

If the only force acting is gravity, then dividing by the mass will give you the acceleration. Gravity is the only source of acceleration, and gravity is 9.8m/s^2.

c) I used t = sqrt (2h/g) 3.15 seconds, + the Orignal t = 5 seconds = 8.15 seconds which is wrong.

Don't know what is wrong here...

It was accelerating for 5 seconds, that doesn't mean it took 5 seconds to get to the top of its flight. Try another approach.

No calculations are needed for b). The acceleration due to gravity is a constant for a given gravitational field.

In part c) 5 seconds is the time till the feul runs out, not the time to the maximum height. Since the initial upward acceleration is consant the velocity when the fuel runs out is 3*5 = 15 m/s use this as vo in v = vo - gt, and solve for t at v = 0. This will give you the time from fuel exhaustion to maximum height.

Still didn't get this can anyone help me with this

wht exactly to substitute...

Plz. help!

Part c)

Upward acceleration of 3m/s^2 for 5 secs. To find the velocity acquired from this acceleration, use the equation

$$v = a t = (3) (5) = 15m/s$$ This is the velocity when the fuel runs out. From here the only force acting is gravity, which is -9.8m/s^2
Find the displacement from this acceleration:
$$x = at^2/2 = (3)(5)^2/2 = 75/2 = 37.5m$$ This is how high it went when it ran out of fuel.

Using the position equation to find how long it takes to fall back down:
$$x = x_0 + v_0t + \frac{gt^2}{2}$$

Total displacement is -37.5m
Initial velocity = +15m/s
Acceleration = -9.8m/s^2

$$0 = -4.9t^2+15t+37.5$$ If you solve this, you'll find the total time of flight from the point when the fuel runs out. Add 5 to this to find the total time. The solutions for t to this equation are

$$t = \frac{2v-\sqrt{v^2+2gh}}{2g}$$ and $$t = \frac{2v+\sqrt{v^2+2gh}}{2g}$$

Notice that your answer for a is incorrect. It travels 37.5m in 5 seconds, but at 5 seconds the acceleration stops. It still has upward momentum, and is still going upwards.

Energy analysis might be easier to solve this problem, but I have class soon :/

Now, can somebody help me on part b,

Part b is the easiest part. It doesn't matter what's happened at any time before or what will happen after. At any time, the force of a body in Earth's gravitational field will be 9.8*m. Whether it was accelerating before or after is irrelevant.

If nothing ihappening, then assume the normal condition that gravity is pulling down.

But how to find m.

F = ma

m = F/a

m = 9.81 / 9.81
So, a = F/m

= 9.81 / 1 = 9.81, Is this right , then

Yes that's right. The only force your considering acting on the rocket is gravity, so

rockets acceleration = g

## What is a solid-fueled rocket?

A solid-fueled rocket is a type of rocket where the fuel and oxidizer are combined into a solid propellant rather than being stored separately. This type of rocket is commonly used for short-range missions and has a simple design, making it more reliable and easier to manufacture.

## How does experimenting with a solid-fueled rocket differ from other types of rockets?

Experimenting with a solid-fueled rocket differs from other types of rockets because the fuel is already mixed and solid, making it more difficult to adjust or change during flight. This means that the design and testing phase is crucial for a successful flight, as any modifications cannot be made during launch.

## What factors are important to consider when analyzing the flight of a solid-fueled rocket?

There are several important factors that must be considered when analyzing the flight of a solid-fueled rocket. These include the weight and distribution of the rocket, the thrust and burn rate of the propellant, the aerodynamics and stability of the rocket, and the environmental conditions during the flight.

## What are some potential challenges when experimenting with a solid-fueled rocket?

Some potential challenges when experimenting with a solid-fueled rocket include ensuring the propellant is mixed correctly and evenly, achieving the desired thrust and burn rate, maintaining stability during flight, and accurately predicting the flight trajectory. There may also be challenges in recovering the rocket after launch and analyzing the data collected during the flight.

## What are some benefits of using a solid-fueled rocket for experiments?

Using a solid-fueled rocket for experiments has several benefits, including its simplicity and reliability, low cost of production, and the ability to achieve high thrust and velocity. It also allows for a shorter development time compared to other types of rockets, making it a popular choice for small-scale experiments and missions.

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