login
? for any physics gurus
-
No new posts
 2006-01-26 4:48 PM |
Elite  2468    Racine, WI |  Subject: ? for any physics gurusThis was a question on a recent Physics quiz: A ball is thrown into the air vertically at a velocity of 16 m/s. How many seconds will elapse until it reaches 18 m/s travelling downward? (Use 10 m/s2 for acceleration due to gravity and neglect air resistance). My answer was that it will hit the ground before it reaches that speed. I got it wrong, with the correct answer supposedly being 3.4 seconds. My instructor basically refused to discuss it with me before I even finished the question. I'm stumped...from what I can figure it will spend 3.2 total seconds in the air and travel to a max height of 12.8 m. By my calculations the ball would have to travel 1.8 seconds downward to go from 0 (when it changes directions ) to 18 m/sec2 and it should hit the ground 1.6 seconds after beginning it's downward fall. It would have to travel 16.2 m to get there also, obviously much further than it's max height of 12.8. Anyone have any thoughts? I googled and found similar questions but nothing quite the same. |
| 2006-01-26 5:18 PM in reply to: #332741 |
Champion 7558  Albuquerque, New Mexico | Subject: RE: ? for any physics gurus Cool! If you make the assumption that you throw the ball up from a flat plane (like a ball field), then you are correct that the ball would hit the ground before achieving 18 m/s. Your calculation is correct. Your professor is also correct. If you toss it up from the roof of a building (holding your hand out so it doesn't land on the roof), then it will take 3.4 seconds. What makes this really cool compared to the first scenario is that it is asymetrical, so you cannot just solve for time to peak height (zero velocity) and double it. Want another really cool problem? Assume you toss the ball straight up from the back of a pickup truck moving horizontally at 20 m/s. Where does the ball land? (Again, neglect air resistance and all of that other stuff.) (This is one of the problems my oldest son had in his HS physics class, and he swore the answer in the book was wrong.) (Yea, I'm a geek.) |
| 2006-01-26 5:49 PM in reply to: #332741 |
Queen BTich 12411 , | Subject: RE: ? for any physics gurus I LOVED physics. Too bad I don't remember much of it. I should take another class just for fun. |
| 2006-01-26 5:51 PM in reply to: #332741 |
Expert 713 WV | Subject: RE: ? for any physics gurus My students have the same issue you had with this type of problem. I will say that at least 50-60% miss this one....even though I go over a similar problem during test review. I am sure this is the equation you used: Flight time = -2v/g. This flight time is figured when take-off and landing are at the same height. However, your prof wanted you to deduce that take-off and landing heights were different. So, the equation used should have been: Flight time = (- v - √(v2- 2gh))/g. I will have to work on the problem McFuzz's son had. Edited by Dr Hammer 2006-01-26 5:51 PM |
| 2006-01-26 5:51 PM in reply to: #332741 |
Expert 713 WV | Subject: RE: ? for any physics gurus Sorry....Fat-fingered the keys. Edited by Dr Hammer 2006-01-26 5:53 PM |
| 2006-01-26 5:58 PM in reply to: #332741 |
Elite 4344 | Subject: RE: ? for any physics gurus Part b of the question would be, "What's the minimum height of the dude that tossed the ball if he were standing on horizontal ground?"
Velocity is -g*t+initialvelocity or -10*t+16. Position is g/2*t^2+initialvelocity*t+initialposition. If final position is 0 ie on the ground then 0=-5*tfinal^2-16*tfinal+initialposition
using the answer from part a tfinal=3.4 seconds
intialposition=3.4*(5*3.4-16)=3.4*(17-16)=3.4 meters
The dude was pretty tall and the prof was just goofing around.
TW
initialposition= |
| 2006-01-26 6:08 PM in reply to: #332741 |
Elite 3650 Laurium, MI | Subject: RE: ? for any physics gurus it's a poorly worded question. The ball would, in fact, not reach the target velocity before passing through the plane of origin (defined as a plane normal to accelerating force). The question didn't say that the ball could fall below the origin, which it would have to do to reach the higher speed. As for McFuzz's Q, I don't feel like working out the math, but just figure out the time it takes for the ball to apex and return to the origin and multiply by the vehical speed to get horizontal displacement. The ball will land back in the truck. Parameterized motion vectors are linerally independant. |
| 2006-01-26 7:59 PM in reply to: #332791 |
Champion 7558 Albuquerque, New Mexico | Subject: RE: ? for any physics gurus vortmax - 2006-01-26 6:08 PM it's a poorly worded question. The ball would, in fact, not reach the target velocity before passing through the plane of origin (defined as a plane normal to accelerating force). The question didn't say that the ball could fall below the origin, which it would have to do to reach the higher speed. As for McFuzz's Q, I don't feel like working out the math, but just figure out the time it takes for the ball to apex and return to the origin and multiply by the vehical speed to get horizontal displacement. The ball will land back in the truck. Parameterized motion vectors are linerally independant. Ding Ding Ding, we have a winner! My son couldn't believe that the ball would land in the truck. |
| 2006-01-26 8:26 PM in reply to: #332741 |
Master 2287 Calgary, Alberta | Subject: RE: ? for any physics gurus Okay you're all too smart for me - or rather it's too many years since my last Physics class. Re the ball hitting the ground before it reaches the speed. Yes throwing it from the side of a building would work. But aren't we forgetting to ask how tall the person is who throws the ball? i.e. it doen't leave ground level at 16 m/s it leaves the persons upstretched arm some height above the ground already. So my question is - how high off the ground would the ball have to be at launch to reach the 18 m/s downward speed before slamming back into the earth. |
| 2006-01-26 8:30 PM in reply to: #332858 |
Elite 4344 | Subject: RE: ? for any physics gurus CalgaryRunner - 2006-01-26 9:26 PM how high off the ground would the ball have to be at launch to reach the 18 m/s downward speed before slamming back into the earth. 3.4 meters. |
| 2006-01-26 10:16 PM in reply to: #332741 |
| Master 1534 | Subject: RE: ? for any physics gurus Great, my physics exam is on monday  . |
| 2006-01-26 10:47 PM in reply to: #332741 |
Elite 2468 Racine, WI | Subject: RE: ? for any physics gurus WOW, thank YOU for all the replies!!!! I feel worlds better knowing that at least my line of thought wasn't completely out of line (pun intended). Not that that helps my grade much...*sigh*...at least I have the rest of the semester to catch it up (my first quiz, then an argument with the prof. It's going to be a loooooong semester!)
|
| 2006-01-26 10:49 PM in reply to: #332761 |
Elite 2468 Racine, WI | Subject: RE: ? for any physics gurus McFuzz - 2006-01-26 5:18 PM Assume you toss the ball straight up from the back of a pickup truck moving horizontally at 20 m/s. Where does the ball land? (Again, neglect air resistance and all of that other stuff.) (This is one of the problems my oldest son had in his HS physics class, and he swore the answer in the book was wrong.) I intentionally didn't look at the answers everyone else posted to this, because when I get to that point in the class I will do the problem (Yea, I'm a geek.) I like geeks. Besides, who but a FELLOW GEEK would have posted this in the first place ?
|
| 2006-01-26 10:54 PM in reply to: #332781 |
Elite 2468 Racine, WI | Subject: RE: ? for any physics gurus Dr Hammer - 2006-01-26 5:51 PM I am sure this is the equation you used: Flight time = -2v/g. You are right this is one, but I used a couple of others to cross check and make sure I wasn't losing my mind. This flight time is figured when take-off and landing are at the same height. However, your prof wanted you to deduce that take-off and landing heights were different. So, the equation used should have been: Flight time = (- v - √(v2- 2gh))/g. hmmm...but he gave us the formulas we were to use, we just had to pick the right one. (It's not really a challenging class,- yet -, which makes this issue all the more frustrating). He wanted us to use vf=vi+at, then calculate times up and down separately. But thank you, I am going to copy and keep those formulas for future reference. I have a feeling they may come in handy down the road |
|
|
|
View profile
Add to friends
Go to training log
Go to race log
Send a message
View album
CONNECT WITH FACEBOOK