Conservation of Energy and projectile motion

[vlmstudent]

I am finishing up a lab write up for conservation of energy, where we launched a projectile into the air and measured the initial velocity and height of the projectile. ( 10 trials)

One of my questions asks if I wanted the ball (projectile) to travel twice as high, how much faster would it have to leave the launcher?

From conservation of energy, I've derived the equation: Ug(initial) = K(final) or mgh (final) = 1/2mv^2 (initial)
When using arbitrary numbers, it looks to me that the initial velocity must increase by approximately 1.41m/s times. Is this correct? If it isn't, is there a way that I can find this out?

When I used this number with my actual data from the experiment, it doesn't equivocate. This might be because my calculations of Ug initial and K final were in disagreement with each other due to extremely small uncertainties. But I just want to make sure that is the only reason why.

Thanks for your help in advance.

 

[berkeman]

I am finishing up a lab write up for conservation of energy, where we launched a projectile into the air and measured the initial velocity and height of the projectile. ( 10 trials)

One of my questions asks if I wanted the ball (projectile) to travel twice as high, how much faster would it have to leave the launcher?

From conservation of energy, I've derived the equation: Ug(initial) = K(final) or mgh (final) = 1/2mv^2 (initial)
When using arbitrary numbers, it looks to me that the initial velocity must increase by approximately 1.41m/s times. Is this correct? If it isn't, is there a way that I can find this out?

When I used this number with my actual data from the experiment, it doesn't equivocate. This might be because my calculations of Ug initial and K final were in disagreement with each other due to extremely small uncertainties. But I just want to make sure that is the only reason why.

Thanks for your help in advance.

Welcome to the PF.

In your future posts in the schoolwork forums, please fill out the Template that you are provided when starting a new schoolwork thread. It helps to organize your post, and makes it easier for us to help you. Thanks.

On your question, it looks like you did the correct thing with the KE and PE equations. Can you post your experimental setup and your data so we can check them?

 

[vlmstudent]

Sorry about that.

I am not as worried about my experimental data adding up because after calculating my uncertainties, I found the values to not be in agreement with each other. I've found a number of reasons for this. However, my particular class is using a lab manual that is still in the editing phases ( yes, we are the guinea pigs), so it was easy to make mistakes here and there.

My main question, is how does the value of initial velocity change if you want to double the height a projectile reaches? Is there an expression I can derive through conservation of energy? This was a projectile launching straight up into the air.

The question on my worksheet is: For the ball to travel twice as high, how much faster would it have to leave the launcher?
Since my data doesn't quite add up, I was planning to use an expression to answer this question instead of my data. If you think looking at my experimental data will help, I have included it as an attachment.

The set up:

A launcher pointing upwards with two photogates measuring time it took for the projectile to pass from the bottom one to the top one. The photogates were approximately 10 cm apart from each other (however I measured them to be 9.5cm). By using this set up I was able to determine initial velocity of each trial. A meter stick was fastened up vertically to measure height of the ball. The 0 of the meter stick was lined up with the bottom photogate. Values for h (y final) were measured by sight only. ( I realize there is room for discrepancy here).

mass of the projectile is not included on my data sheet. It is 0.016116 kg

Issues with data:
I reported height (h) to nearest cm. (values are reported in meters), while it may have been in my favor to attempt to calculate to the mm. By eyesight, this is rather difficult. But the experiment didn't specify this.
Also... the distance from the top and bottom photogate was different from one side to another. I should have measured the other side that was braced to a bar, which would have likely given me a measurement closer to the 10 cm I was looking for. Thanks again for your help!

 

[berkeman]

Sorry about that.

No worries

I am not as worried about my experimental data adding up because after calculating my uncertainties, I found the values to not be in agreement with each other.

Your data look pretty close.

$$\frac{1}{2}mv^2 = mgh$$
$$(3m/s)^2 = 2 * 9.8\frac{m}{s^2} 0.5m$$
$$9\frac{m^2}{s^2} = 9.8 \frac{m^2}{s^2}$$

Certainly determining the launch velocity with sensors that far apart is problematic. Which direction would that give you an error in your data? Is it in the right direction to help to explain the difference in the two numbers in my last equation?

By using this set up I was able to determine initial velocity of each trial.

Or did you already apply the correction needed to account for the separation of the 2 gates? If not, can you say how you could calculate how to correct that time to give you the real initial velocity at the first photogate?