I can't figure out what I'm doing wrong in finding position. If anyone can help I'd appreciate it.
#1.) If a ball was thrown staight up into the air at 20 ft/s, find the position and velocity after 1.2 secods.
I've found the Vf by Vf=Vo+at which is -18.64ft/s
but I can't figure out what equation to use, or steps to follow, to find the position.. I've tried both
Y=1/2(Vo+Vf)t and Y=Vot+1/2g(t^2) but neither come up with the correct answer of 4.68ft
also #2.) a golf ball leaves the club at 124 ft/s at an angle of 50 degrees above the horizontal. What is the range?
I've used the formula R= (Vo^2 x sin(2theta))/g and i get 470.26 ft which is double the answer (235.1 ft). should the gravity be halved for some reason? Isn't acceleration by gravity -32.2 ft/s?
I just can't figure these two out. I can do compound vectors but I just can't figure out this stuff. Any help would be greatly appreciated.
WTF ... you have to break it up ... Your velocity is right it looks like though ...
20ft/sec will be done in under 1 second. Gravity is 32 feet /sec/sec. ...
step 1. V = u + at ... so to a stop, v = 0 and g is negative so 20/32 = t. = .625.
Then you have .575 seconds of gravity. so your velocity is .575X32 downward.
Ironically it will pass that same point in space as it did on the way up with the same velocity in the same time.
So it will be exactly where it was on the way up at that speed.
So v2 - u2 = 2as.
400 - 338.56 = 2 X 32 X distance. = .96 ft.
Yea its about to clobber you at 20 clicks and its .96 away ... whatever ...
Cool.
Srinath.
Parabolic stuff, I split it into vertical and horizontal velocities, use the vertical to calculate when it will come to a stop in the vertical, double that time and multiply it with the horizontal for range.
Cool.
Srinath.
Nerds :flipoff: :flipoff: My take on the Subject what Happens Happens when it Happens. :thumb: :thumb:
thx for the help but I figured out my problem after I the professor emailed me back after I showed my work, he said he must have gotten a wrong calculation and his answers were actually the ones that were wrong. :mad: seriously... I spent like 3 hours trying to figure out my mistakes and dwelling on this just to find out he made the mistake.. :mad:
Grrr. That sucks. It happen, though. Welcome to academia.
Also, drop the class. Then find a professor who uses real units to do physics in. :icon_mrgreen:
Dont ween on the nipple of the SI system...ft/BTU ect..ect.. are used everywhere in industry still and unless you want to spend double the amount of time on problem converting learn to use them.
I agree that the english units suck big ones...but they are used to commonly
Yup ... they screw up all the time.
BTW what was the right answers.
I dont blindly apply formulae ... I always break them all up into little pieces.
Cool.
Srinath.
Kenyon to Professor wrote:
I still am not producing the correct answer, here is my work:
#2.) To find Vf:
Vf = Vinit+gt
Vf = 20ft/s + (-32.2ft/s ² • 1.2s)
Vf = 20ft/s - 38.64ft/s
Vf = -18.64 ft/s
To find Ymax
Vf² = Vinit²+2gy
0 = (20ft/s)²+2(-32.2ft/s ²)y
0 = 400ft²/s² - 64.4ft/s²y
64.4ft/s²y = 400ft²/s ²
y = 400ft²/s² / 64.4ft/s²
Ymax = 6.21ft
then to find Ydown
Vf² = Vinit²+2gy
(-18.64ft/s)² = 0+2(32.2ft/s²)y
347.45ft²/s² = 64.4ft/s²y
y = 347.45ft²/s² / 64.4ft/s²
Ydown = 5.395 ft
Then Ymax 6.21ft - Ydown 5.395 ft = position of .82 ft above the starting point.
even if I do it with both Vf and Vinit I get the same answer
Vf² = Vinit²+2gy
(-18.64ft/s)² = (20ft/s)² + 2(-32.2ft/s²)y
347.45ft ²/s² = 400ft²/s² - 64.4ft/s²y
347.45ft²/s ² - 400ft²/s² = -64.4ft/s²y
-52.55ft² /s² = -64.4ft/s ²y
y = -52.55ft² /s² / -64.4 ft/s²
Y = .82 ft
#5.) A golf ball leaves the club at 124 ft/s, 30° above the horizontal. What is its range?
R=(Vo²•sin(2Θ))/ gravity =
R=(124ft/s²•sin(2•50°)) / 32.2ft/s =
R=(15376 • .9848) / 32.2ft/s
R=15142.404 / 32.2
R=470.26 ft
but the given answer is 235.1 ft? Is range only the distance to Ymax or fron Xinit to Xfin?
Could you help me find my mistake? Am I using the wrong formulas or using them incorrectly?
From professor to me:
You are right, my answers were wrong. Hope that helps
Good Luck
I don't really care about the systems or blindly applying formulas b/c I probably will never use this stuff ever again after I'm out of college. I mean, I will be able to do it again with looking up formulas and examples but I can't see any real world applications in my line of work.
Quote from: ixolas on February 17, 2007, 03:41:47 PM
From professor to me:
You are right, my answers were wrong. Hope that helps
Good Luck
:laugh: :laugh: Thats a nice kick in the face :mad:
Ya.. now my confidence has dropped a little. So now i'm going to look closely at the tests and if I miss any I'm going to have him point out my mistake so he doesn't have any more of these "oh my bad".
Quote from: ixolas on February 18, 2007, 12:52:23 AM
Ya.. now my confidence has dropped a little. So now i'm going to look closely at the tests and if I miss any I'm going to have him point out my mistake so he doesn't have any more of these "oh my bad".
yeah that happened in my physics class once. right after a test almost the entire class got the same answer and the prof was bitching about how retarded we all were, not knowing how to do simple grade school calculus and all. So we got up did the work on the board and proved he was indeed a jackass. He bought us pizza and blamed is TA :laugh: :laugh: :laugh:
Ya its always someone or something else's fault.. well at least he bought pizza, that was cool of him.