Initial Velocity Question

[Cyrus]

What supplies, you have an equation!

Write the equation down for me! We just took care of all the terms!

Come on now, you know this. You were the one that supplied me with all the information.

 

[MaNiFeST]

0 = Voy^2 + 2(-9.81m/s^2)(1m)
19.62=Voy^2
4.429m/s=Voy

! I just made a = -9.81 for going up so wouldn't end up with negative answer

 

[Cyrus]

Yes, that is correct, in this equation you must make the acceleration -9.81m/s, do you know why?

 

[MaNiFeST]

So if the Voy is 4.429m for 1 m , then wouldn't it be different if the displacement was bigger? so how would i know how many times to multiply 4.429 by?

SO

4.429m/s = Vo * sin(90)
Vo = 4.429 m/s

 

[MaNiFeST]

I must make a = -9.81 because it is going away and cannot have a negative velocity?

 

[Cyrus]

Whoa, sir. Why are we back to introducing this equation?

We just used the only equation of importance here, namely: $$V_{f}^2=V_{0}^2+2a(ds)$$

 

[MaNiFeST]

Yeah sorry, i forgot that sin(90) is 1 , so they are the same because it is going vertical anyway

 

[Cyrus]

I must make a = -9.81 because it is going away and cannot have a negative velocity?

What is the effect of gravity on the balloon as it moves up?

 

[MaNiFeST]

gravity is pulling it down to earth, but you can use it either way + or - , you just have to change it to - when the balloon comes back down

 

[Cyrus]

It's not an 'either way', it depends on the physics of the situation. Is it going up? If it is, then gravity acts to slow it down and acceleration must be negative, or deceleration.

If its going down? Then gravity is acting to accelerate the body and it is positive.

This is not just a pick which way gives you the write answer.

 

[MaNiFeST]

oh , well in my physics class atm we arnt dealing with deceleration, only one form of acceleration , but yeah i get what you mean

Thanks again

 

[Cyrus]

It is the same thing, except there is a minus sign for deceleration.

 

[Cyrus]

So if the Voy is 4.429m for 1 m , then wouldn't it be different if the displacement was bigger? so how would i know how many times to multiply 4.429 by?

SO

4.429m/s = Vo * sin(90)
Vo = 4.429 m/s

This is not correct, the change in initial velocty is not a linear change with stop height.

I.e., if it goes 1 meter and 4.429m/s, it will NOT go 2meters for 2*4.429m/s, because the velocity is squared:

$$V_f^2 = 2a(ds)$$

Plug in numbers for different heights, ds, and post your results.

Hint: Choose multiple values of change in ds, or height. I.e., like 1m, 2m, 3m,4m etc. etc. etc.

 

[MaNiFeST]

5m = 9.9045
7m = 11.719
10m = 14.007
12m = 15.344
15m = 17.155
20m = 19.809

 

[Cyrus]

Good, now compare 5 and 10 and 15 and 20. What do you notice?

 

[MaNiFeST]

its about
1.12
2.25
3.38
4.51

 

[Cyrus]

so, fix what you said earlier:

So if the Voy is 4.429m for 1 m , then wouldn't it be different if the displacement was bigger? so how would i know how many times to multiply 4.429 by?

 

[MaNiFeST]

Hmm i dunno, how would i measure a displacement going higher than 1m? i don't think i can exactly mark the walls :(
there must be a way
sorry lol i just really want that extra credit :)

 

[Cyrus]

Do not worry about measuring it right now, I am trying to get you to learn about the physics. Fix that statement above.

Hint: Look at your ratio. Your height changed by a factor of 2 from 5 to 10 meters, how did your velocity change?

 

[MaNiFeST]

ohhh
1.12*2 = 2.24
2.25*1.5 = 3.375
3.38*1.33 = 2.541

 

[Cyrus]

Not quite... What I wanted you to do, although a little bit hard to spot, is notice that when you DOUBLE the final stop height from 5m to 10m, the velocity increases NOT by a factor of 2, but by a factor of $$\sqrt(2)$$.

Do you see this?

 

[MaNiFeST]

sorry i don't really follow

 

[Cyrus]

Look at the change in height.

10m/5m = 2 <-- you DOUBLED the height it stopped at

14.007/9.9045= 1.4142 <--does NOT, does NOT, does NOT DOUBLE.

1.4142 = sqrt(2) <--it changes by the square ROOT of the change in height.

 

[Cyrus]

Try it for a different change in heights and see if it works.

 

[MaNiFeST]

15/5 = 3
9904*sqrt(3) = 17.154
It works!
Thanks dude

 

[Cyrus]

Go back over everything we did, print it out, and show it to your teacher.

You will get your bonus. You solved this on your own.

 

[MaNiFeST]

I didnt really do this on my own, but thanks dude, you really helped me!

Yeah i do understand this, ill have to remember last part, but tricky

 

[Cyrus]

Yes, go back over this tomorrow and make sure you understand every step. If you dont, ask me more questions until you UNDERSTAND it.

 

[MaNiFeST]

Hey, i found out that we can use a protractor on the lab. I read some where that you can use a proctractor to find out the distance or height of something from ground level.

I tried searching google but didnt really find anything, do you have a helpful tutorial or maybe some tips on using a protractor to determine the height of something launched 90 degress into the air?

 

[drpizza]

I do a lab with a water balloon slingshot with my students... I'm sorry, but there's no realistic way you're going to measure the vertical displacement with a meter stick... There's nothing nearby for reference; the water balloon goes well above the height of the school and any other vertical reference point. It's not much different than suggesting to someone that they can measure the vertical height a model rocket reaches by using a meter stick. (well, a rocket usually goes much higher, but regardless, you're well out of the range where using a meter stick would be a reasonable suggestion.)

Manifest, I was going to suggest to you that you could stand at a distance from the waterballoon slingshot, then aim the meter stick toward the highest point the balloon reaches. Constructing a smaller triangle from where the meter stick is pointing, you could use similar triangles to calculate the maximum height of the waterballoon. However, now that you're allowed to use a protractor, right triangle trig simplifies the work you're going to have to do (slightly).

btw, I appreciate this thread... after calculating an initial velocity, my students then have to calculate a horizontal range after pulling an angle out of a hat. (I stand at their calculate point, getting soaked if all goes well.) Students are notoriously horrible at getting accurate times with a stopwatch.

You'll definitely want to use the formula that cyrusabdollahi led you to. However, I agree with your skepticism toward getting credit for suggesting that you measure the vertical displacement with a meter stick.