Rocket Lab, Finding Initial Velocity, Height Given time.

1. Oct 1, 2008

Hellsing834

1. The problem statement, all variables and given/known data

We are doing the Rocket lab in my Physics class, we shot rockets with different caps on them [ Low, Med, High ] and recorded the times. I have the time, i need to find Initial velocity and Height given that i have time. I also need to take into consideration air resistance [ Not sure, but i think i heard my teacher say that]

2. Relevant equations

I was thinking that if i did V = Accel*time, that would give me the velocity, but thinking that i need to include air resistance, would i need to start doing free body diagrams?

3. The attempt at a solution

If i did the diagrams that the net equations would be :

Fx = 0
Fy = n-g. [ Normal - Gravity]

I cannot think of any other factors to include, but the problem is i don't know how what to do now. I am stuck at the Fy equation.

Also, if there was no air resistance included, would i just do V = a*t to find my initial?

2. Oct 1, 2008

GTrax

Just to be clear about how the rockets got "shot" into the air. The rockets were not burning up mass during the boost upwards, like a fireworks rocket. Instead, the entire flight was passive, as if they had suddenly been thrown up with an initial velocity, like a stone - right?

If so, then the acceleration downwards was g = 9.81m/sec^2. Use the equation of motion v=u+at.
In this case, the "a" is replaced by -9.81. It has a minus sign because we decide that up is positive.

You can do it several ways. The velocity was zero at the peak height. It took the same time to get up there as it took to fall back. Once you know the time, there is only one answer for the rest. You get the initial velocity.

The height also comes out if you know the time to get to the top, where the velocity was zero.
Learn the equation for distance covered. For steady velocities, distance = u*t where u is the velocity, and t is the time.
Where it is modified by an acceleration, distance = u*t + (1/2)*a*(T^2). In this case, a = -g. Get the SIGNS right.

Now - a key point about these accelerations. IF this rocket was burning to provide a steady force F, and it was opposed by gravity, then the net acceleration up would be (F/m)-g. The trip down would be a straightforward drop with acceleration = -g. It becomes a 2-part problem.

That should be enough tools to do the job. Try it.

Last edited: Oct 1, 2008
3. Oct 1, 2008

Hellsing834

Well what we did is, The rockets were air pumped. so after a certain amount of pressure, they shot up in the air.

Now how we recorded time is that when the rocket reached its maximum height, we would start timing, and when it hit the ground, we would stop the timer.

also, the equation v=u+at, what does the u stand for? also, would have any idea on how to incorporate air resistance in there?

4. Oct 1, 2008

GTrax

u is the initial velocity
eg. something takes 5 seconds to get up there

v = u + at becomes 0 = u - 9.81* 5

We set v = 0 meaning thats where it came to a halt at the top.

The initial velocity here would have been 49.05 m/s upward

5. Oct 1, 2008

GTrax

Importantly, it takes the same time to get up there as it spends falling back - unless it was burning an engine to keep giving it a shove.

Can the release of air pressure be considered to be a sudden hit, where the rocket is suddenly given an initial velocity - or does the air pressure exit keep up the force for a significant time during ascent?

6. Oct 1, 2008

Hellsing834

yea, thats why we only timed the half. I get that, and i can even figure out the distance, but i am not sure how i can incorporate air resistance in there. I realize that i can subtract the Weight of the rocket from the velocity? That would take into account the resistance, but weight = mass x gravity, and we don't have mass..or weight.

7. Oct 1, 2008

GTrax

Air resistance is a function of velocity. It will have a force F(air) = K * v where K is some constant that will depend on the rocket shape and size.

eg. a skydiver will fall with acceleration g, but from this is subtracted the effect of air resistance. A way to figure it is to wait until he stops accelerating. At this point, he has reached terminal velocity - in effect simply continuing to move in a state of uniform motion, because the gravity force m*g exactly equals the air resistance force K*v

Note that the acceleration of things rising and falling, affected by gravity is the same, no matter what the mass.

Not so for the acceleration component introduced by air resistance, or air pressure shoves, or rocket motors, or anything else whatever. In those cases, the mass is fundamental to how fast they get moving.

That is why I think the air resistance is neglected in this experiment - unless you noticed the rocket reach a steady speed while falling.

8. Oct 1, 2008

GTrax

BTW - you cannot subtract a weight from a velocity. A weight is m*g force. A velocity is distance per second. You can add and subtract FORCES. You can add and subtract accelerations, remembering that you need the mass to figure an acceleration from a force. In these calculations, it need not appear.

How long did it take to get up there?

9. Oct 7, 2009

jodyhannah

in the equation above, what does the "T" stand for? i get everything else, but not that one part...

10. Oct 8, 2009

GTrax

Hi jodyhannah. This is a very old thread from 2008. Anyways, t normally represents time when used in equations of motion. Clearly, in the equations above, a capital T was inadvertently typed in the equation distance = u*t + (1/2)*a*(T^2). In that equation, the T needs to be lower case. Very sorry about that. I hope you are OK with the equation now.