Calculating Water Rocket Height from Hangtime | Formula and Explanation

[Justin Murphy]

Homework Statement

We are doing water rockets and I don't know how to calculate the height it went from it's hang time. It's total hang time was 7.35 seconds, and I know it's vertical acceleration was -9.8m/s^2 from gravity, and I'm assuming I need to first find out it's initial velocity, since I assumed the equation I would be using is x=V0t+1/2at^2

Homework Equations

x=V0t+1/2at^2

The Attempt at a Solution

I don't know how I got it, but a random calculation got me around 75ish meters, but I don't know what equations or variables I used (that was yesterday.)

 

[scottdave]

So if you ignore air resistance, then you can assume that half of the time is spent going up, and half of the time is spent going down.
Use this formula to find the initial vertical velocity: V_final = V_initial + a*t. What will be V_final when the rocket is at the peak?

 

[Justin Murphy]

The final velocity at the peak would be zero. How would I switch around that equation to solve for the initial velocity?

 

[Justin Murphy]

I solved for V_initial and got 14.3325m/s.
1. Is this correct?
and 2. does that mean it's total height is 232.75m? (I plugged it into the distance equation, and I'm no expert but this seems a little high)

 

[haruspex]

I'm assuming I need to first find out it's initial velocity,

You don't. There are five standard SUVAT variables. Any four are linked by an equation, so there are five standard equations. You know time, acceleration and final (i.e. at highest point) velocity, and you want the distance. Pick the equation that involves those four.

solved for Vinitial and got 14.3325m/s.

Too low. Please post your working.

232.75m? (I plugged it into the distance equation, and I'm no expert but this seems a little high)

It is too high.

 

[Justin Murphy]

Ok, I asked someone in my physics class and they gave me the equation d=1/2at² and when I plugged it all in I got about 66 meters. NOW is it correct?

 

[haruspex]

Ok, I asked someone in my physics class and they gave me the equation d=1/2at² and when I plugged it all in I got about 66 meters. NOW is it correct?

Before I answer that, explain why that equation is appropriate.

 

[Justin Murphy]

I'm not completely sure, but I know part of it is because I'm solving for distance, and part of it is because I have the variables to plug in. Our teacher gave us this equation in class, and I think I forgot to write it down because I couldn't find it in my notes, and then it turns out to be the one we use the most. From what I know, it's reduced from that first equation from my original post, and I actually had the right equation the first time, but I was just solving it the wrong way.

 

[scottdave]

I'm not completely sure, but I know part of it is because I'm solving for distance, and part of it is because I have the variables to plug in. Our teacher gave us this equation in class...

This would be a good time to review notes or read the book (if your class uses one) or look at some videos on the subject. It is important that you understand what is going on, so you will understand which formula is applicable in the situation.

 

[Justin Murphy]

Thanks for the help

 

[scottdave]

You could start here: https://en.wikipedia.org/wiki/Equat...translational_acceleration_in_a_straight_line
You can memorize them all, but I only have the first two committed to memory, then derive the others, as needed. So yes, in my case I would first figure out v₀, then go from there.

Also, make sure you are using the correct value for time. If you are looking at the time from start to the peak height, that is going to be less 7.35 seconds, if I understand that "hang time" refers to total time in the air. If you state assumptions (such as air resistance is negligible) then you should say that then say that you assume half the time to go up, and half to go down.

 

[Justin Murphy]

Yes, the hang time was the total flight from launch til it hit the ground, and I made sure to cut the time in half in the equation. My only problem is the amount of formulas, because I don't have very good memory recall, and I'm not the best at reworking base equations to suit my problem.

 

[scottdave]

I solved for V_initial and got 14.3325m/s...

OK let's start with that. The initial velocity is not 14.3 m/s. But can you show the steps that you took, to arrive at this. We will help you find your mistake, so you can get the correct answer.

 

[haruspex]

it's reduced from that first equation from my original post,

Yes, but it's not a good idea to have a proliferation of equations to remember for special circumstances. Better just to stick with the standard five SUVAT equations and get used to applying them.
In the present case, the Relevant Equation in post #1 was exactly the one to use. The trick in applying it was to consider only the second half of the trajectory, as the object falls to ground from its highest point.

I also suggest you take up scottdave's offer to find where your calculations went wrong.

 

[Justin Murphy]

Part of my problem was when I used the total time instead of half the time.