anonymous
  • anonymous
If a rocket is launched at a 30 degree angle, with a velocity of 36.7 m/s, an average time of 4.3025 s, and a distance of 116.63 m, what is the highest height it reaches in the air?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Average time of what exactly? And what's that distance supposed to be? Your question doesn't make any sense - please restate it more accurately.
anonymous
  • anonymous
Average time the rocket is in the air. The distance is how far the rocket traveled. I need delta y max, or the height at the highest point the rocket reaches while in the air.
anonymous
  • anonymous
So your rocket is not actually a rocket but is ahm.. "thrown" ??? Or do you have numbers on how much fuel is burned and what the exhaust velocity is etc? Distance on the ground?

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anonymous
  • anonymous
Yes, distance on the ground. And yes, sorry, the rocket is launched from a tennis ball launcher, so thrown is correct. This was an activity we actually did in physics class, so these are real numbers.
anonymous
  • anonymous
Okay, just for the future - not just here online - try to be really specific and precise, when it comes to ahm.. science and stuff ^_^ Also: Where exactly are you stuck at? Have you gotten anywhere on your own, yet?
anonymous
  • anonymous
Yeah, sorry about that, I guess I forgot about those details. I mainly need an equation that applies to this situation. Or is it using trig? That's mostly it--I'm not really sure what I should be using to solve it.
anonymous
  • anonymous
Well ahm.. there's really more than one way to do it. For starters: Do you think that - in general - the time an object stays in the air has anything to do with how far it moves horizontally?
anonymous
  • anonymous
Well...we got an equation [delta x = V (sub x) * delta t] where delta x was the distance covered and delta t was the change in time. That's what I assumed it was talking about.
anonymous
  • anonymous
Well ahm.. that equation doesn't really have to do with what you want to know at all.. so ahm.. about my previous question - what do you think? |dw:1354261442965:dw| If you were to throw 2 objects like the poor drawing I made. Would one object be longer in the air if they both reach the same height? (we are disregarding friction here, but that's ok)
anonymous
  • anonymous
Well, the one with the further distance would be in the air longer, correct?
anonymous
  • anonymous
Nope, I'm sorry. In fact, it does not matter how far you throw an object at all - it only matters how "far" you throw it vertically - how high. I you would get on the roof of your house and let a tennisball fall straight down and at the same time throw one horizontally, they hit the ground at the exact same time. The "vertical component" of their movement is exactly the same. They both get accelerated at 9.81 meters per second per second - _every_ object on earth does. Thus, every object on earth falls at the exact same rate. (In reality, if you got something like a snowflake, there's so much air-friction, it does slow down, but calculating air-friction is rly complicated and normally you just omit it and it's ok. I'm sure, your teacher doesn't want you to take it into account.)
anonymous
  • anonymous
Once, you understand that, you see that we would only need the time it takes your rocket to hit the ground, to calculate how high it went. But as we are even given it's initial speed and the angel at which it's thrown, we could also - for example - apply trigonometry and figure out it's vertical velocity at the beginning.. which way do you want to do it? (that's not the only possibilities btw..)
anonymous
  • anonymous
Oh! Right! Yeah, I remember learning that now. Um, how complex is the equation that would use initial speed and the angle?
anonymous
  • anonymous
Well - it's not complicated at all. (maybe you'd disagree ;) ). Anywho.. I'd like to hear it from you - I am sure, you can figure this out... What do you know about the relationship about velocity and acceleration and all that stuff?
anonymous
  • anonymous
Ok, wait, is this [delta y = V initial y *t + (1/2) gt^2]?
anonymous
  • anonymous
Thats somewhat correct - depends on what your value for g is. Cause gravity is going to accelerate the rocket to the ground - that means, in the opposite direction of your initial speed. Therefor your g has to be negative, or - if it's positive - it's gotta read \[y(t) = - {1 \over 2}g t^2 + v_0 t + y_0\](mind the minus sign in front of the acceleration term). y_0 is just your y-component at the start. This can just be 0, if you are launching from the ground.
anonymous
  • anonymous
Oh yeah - and as you correctly stated, that v_0 is supposed to be the y-component of the speed.. we have to calculate what it is with simple trigonometry. Also - _when_ do you think, your rocket will be at the highest point? At what time?
anonymous
  • anonymous
Yeah, we were told g= -9.81, or about. So wait, is y_0 = 0 or do we calculate it? And it'd be halfway, or 2.15125 s, right?
anonymous
  • anonymous
That's correct - the rocket will be at it's highest point exactly "halfways" through the air. And yes, y_0 is zero, if you launch from the point ground. I just wrote the equation like this, so you could use it even if you launched from like 2 meters above ground or whatnot. (Although then, the fact with t/2 being the highest point would no longer be true). In any case: all that's left to do is to figure out v_0 in the y-direction and plug it in ...
anonymous
  • anonymous
Ok, so if we're using trig does that mean this involves a triangle? Cause how would the velocity apply to that?
anonymous
  • anonymous
Hmm.. do you know about sin() and cos() and the relationship between hypotenuse and catheti and angles and stuff? (imma try to draw something again ^^)
anonymous
  • anonymous
yeah, i know the trig functions.
anonymous
  • anonymous
|dw:1354264081889:dw|So v is the hypotenuse of our triangle and v_x and v_y are the catheti. Now you just gotta figure out how v_y relates to alpha and v. It's really not hard - just give it a try ;)
anonymous
  • anonymous
So 30 = sin(V), then solve for V?
anonymous
  • anonymous
no, wait!
anonymous
  • anonymous
I mean sin(30)=V_y/V.
anonymous
  • anonymous
That's correct! Very good! :)
anonymous
  • anonymous
So is V the velocity given originally? 36.7 m/s? And then is V_y the height? Or do I have that screwed up?
anonymous
  • anonymous
Well yes, v is the velocity, you were given initially. v_y is the 'y-component' (vertical) of this initial velocity. Since the horizontal component id not needed for our calculation, we do not need to get it, but it would be similar:\[v_y = \sin(\alpha) v\]\[v_x = \cos(\alpha) v\]Now what do you get for v_y?
anonymous
  • anonymous
Approximately -36.261 m.
anonymous
  • anonymous
First of all: v is a velocity. That means, it's distance over time, thus it's unit gotta be meters per second (or something similar). Be sure to pay attention to this, whenever you do physics-problems. Messing up ur units can get you in loads of trouble. 2nd: That's wrong, I'm sorry. You should see, it's gotta be wrong. How can the velocity in the vertical direction be negative? And all of it is like 36.7 meters per secon. If you shoot it at 30° wouldn't you expect something like _roughly_ half of it being in the y and half in the x?? Hint: When using trig-functions, always pay attention to this degree/radiant-stuff. Do _not_ enter your angle in degree, when your calculator is set to radians ;)
anonymous
  • anonymous
That's what it was--my calculator is set in radians from the calc test I took today :P
anonymous
  • anonymous
Ok! Much better: 18.35 m/s. I was a bit confused when sin(30) wasn't a nice, even number.
anonymous
  • anonymous
That is correct. Now you just gotta go back and plug everything in ;)
anonymous
  • anonymous
Ok, I got 169.750 m.
anonymous
  • anonymous
Hmm.. just think about that result. Doesn't 170m sound a liiiittle bit high for you? 1. remember t = 2.15 for our highest point, not 4.3 2. be sure to use the right sign on your acceleration ;)
anonymous
  • anonymous
Ok, is the value of g that's being used here -9.81?
anonymous
  • anonymous
As I mentioned before, that depends on your sign for the acceleration term. Just make sure, the resulting acceleration is negative. (cause our y-axis is going upwards, not downwards)
anonymous
  • anonymous
Alright, what I got was 62.126.
anonymous
  • anonymous
Hmm, you plugged in the right time, yet your sign for the acceleration term is still wrong. Again: If you are going to use a positive g, you need to use\[y(t) = -0.5 g t^2 + v t\]If you are using a negative g, you need to use\[y(t) = 0.5 g t^2 + v t \]Your acceleration has to point inthe negative y-direction. (This is, cause we are using a y axis that is going upwards. Thus our initial velocity upwards is also positive.)
anonymous
  • anonymous
Ok, maybe this time: 16.779
anonymous
  • anonymous
...I take it it's wrong :P
anonymous
  • anonymous
If that's meters, it looks good to me ;) One last thing I'll briefly mention is: If those numbers really come from an experiment, the time you are given is pretty ridiculous: No way in HELL can you measure that time to 1/10000 of a second. If you did it "by hand" you probably got an error of about 0.1 - 0.2 seconds (if not more). So using a t-value with more than one decimal place is just pure bullpellet and gives the impression of precision that's simply not there. (That's not that important right now, but if you gonna continue to do physics at a college-lvl or whatever, you should pay attention to it ;) )
anonymous
  • anonymous
Oh, we took an average of 4 different times that we measured with stopwatches (: That's how we got that. But noted! And thank you SO MUCH for all your help! It's very much appreciated.
anonymous
  • anonymous
I figured, you did that ;) It's just - I'm tutoring a physics-lab at my university for example and that's the one thing, were I reaally get mad, when students do that.. that's why I wanted to mention it - it's just baloney. But nevermind - you don't have to wrack your brain with that (yet). Glad I could be of some help - you are very welcome :)
anonymous
  • anonymous
Got it (: Well, thanks again, and good-bye!

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