|
Sam heaves a shot with weight 16-lb straight upward, giving
it a constant upward acceleration from rest of 45.7 m/s2
for a height -----> 69.0 cm<----( Only goes up 69 cm! Doesn't go up higher than this point! ) He releases it at height 2.30 m above the ground. *Meaning the ball released his hand @ 2.30m above the ground* You may ignore air resistance.
What is the speed of the shot when he releases it?
How high above the ground does it go?
How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.82 m above the ground?
It's not really a homework problem; she just posted it on the board and said try it at home , see if you can solve it ; and i wanna be the cool kid that knows the answer 
Of course, I'm not blindly giving you this question without working on it, however i probably messed everything up :D
I don't expect anyone to do it but... if you could point me in the general direction that would be pretty awesome [: .
;edit;
Excuse my godawful drawing T_T
;edit2; Also 69cm = (69)1x10^-2 m or .69 m
And wow TL is really helpful :DD
Also, i suppose that when the ball is coming down ; g ( acceleration), due to gravity = ~9.81m/s^2
Ahh and also this wasn't a homework problem or anything she just gave it to us as a kind of eye opener , she's going to intro./ review it tomorrow in class.
So far we've just been doing physics I review so we're starting kinematics, plus the hurricane set us back a week( no school) so that would explain the lack of movement in the class? D:
Thanks for your help again !
 
|
omg lol i did this but i dont remember it, almost exact problem too, it was a rock. O____O physics 2 seem easier than 1. =D
|
Vf=Vi+at
D=Do+Vit+1/2at^2
Vf^2=Vi^2+2ad
Those are the formulas you need to use, and you should be able to figure it out.
|
Give me a bit, i'll post my answer in a spoiler. work through it yourself and check your answer to mine.
EDIT: @nevake: hahaha, I went through work and kinetic energy since i couldn't remember the formulas. oh well.
+ Show Spoiler [Velocity (Process)] +a=45.7 m/s^2 d=1.61 m (2.30 m - .69 m) m=7.25 kg (converted from lbs, screw english system.) F=ma (Newton's 2nd) W=Fd (work=force times distance force is applied) W=K(final)-K(initial) (work= difference in kinetic energy) K=1/2*mv^2 That's all the formula's you'll need. If you haven't covered all of those yet, well, that's why your teacher said to take it home and think about it instead of assigning it for homework. Time to start plugging in. F=ma W=mad K(final)=mad (since initial K is 0, just drop it.) 1/2*mv^2=mad (mass cancels on both sides, do a bit of algebra, and...) v=sqrt(2ad) Well, okay, now that I think about it, you probably know a kinematic equation that looks like that. It's been a year or two since i've taken physics, gimme a break!
+ Show Spoiler [Velocity (Answer)] +v=sqrt(2ad)
v=sqrt(2(45.7 m/s^2)(1.61 m))
v=12.1 m/s
+ Show Spoiler [Height above ground (process)] +
Yay conservation of energy. Makes this problem cake. If I skip a few steps, sorry.
U=K (conservation of energy)
U=mgh
mgh=1/2*mv^2 (mass cancels)
h=v^2/2g (don't forget to add 2.3 m to this because i'm assuming h=0 where he let's go of the ball)
+ Show Spoiler [Height above ground (answer)] +
h=(v^2/2g) + 2.30 m
h=9.77 m
+ Show Spoiler [Time (process)] +Since I did this a sort of strange way, I'll spare you the details and just give you the basics. So we already know the velocity of the ball when he launched it, and the max height. What I did, was figure out the velocity of the ball when it was at height=1.82 m coming down. To do this, figure it at v=0 at the peak, figure out the distance from the peak to 1.83 m and use v=sqrt(2ad) from before to get velocity. I came out with 10.5 m/s. This gives you a total change in velocity of 22.6 m/s, and the acceleration of gravity is 9.8 m/s^2. a=dv/dt (pretend the d's are delta), and grab a calculator. I got:
+ Show Spoiler [Time (answer)] +
|
It's not really that hard, although it's annoying that she gives it to you with such strange values (mixing english and metric is especially annoying). Just realize that you have enough information to calculate the starting kinetic energy, which is equal to the overall change in potential energy (mass times gravity times change in height).
|
On September 24 2008 06:30 nevake wrote:
Vf=Vi+at
D=Do+Vit+1/2at^2
Vf^2=Vi^2+2ad
Those are the formulas you need to use, and you should be able to figure it out.
Thank you. Can you help me derive the kinematic formulas by any chance? 
|
On September 24 2008 06:30 nevake wrote:
Vf=Vi+at
D=Do+Vit+1/2at^2
Vf^2=Vi^2+2ad
Those are the formulas you need to use, and you should be able to figure it out.
It always bugs me when people answer physics questions like this. If you're just blindly combining formulas from some formula sheet then, first of all you're not learning anything, and second there's a good chance that you'll do it wrong.
|
I was under the impression that he had not learned any of this, and so by giving him these formulas, I was hoping to let him figure out the answer by himself. And your physics textbook should show you how to derive these formulas.
|
|
|
If he's already learned it then did he not learn the formulas? If he had really learned he would already know which formulas to use because it would be intuitive. When you just copy paste formulas like that it just looks like gibberish, anyway. You didn't even define your variables.
|
On September 24 2008 06:38 nevake wrote:
I was under the impression that he had not learned any of this, and so by giving him these formulas, I was hoping to let him figure out the answer by himself. And your physics textbook should show you how to derive these formulas.
there were no more physics C textbooks here and due to hurricane ike , we won't have them for a while T.T
However i googled some stuff and i should ... figure i out..
i hope
WB SHALLOW ^_^ how goes the army life
|
LOL are you serious? Your school doesn't have any textbooks? That's so ghetto.
|
On September 24 2008 06:44 Luddite wrote:
LOL are you serious? Your school doesn't have any textbooks? That's so ghetto.
There's only one Calculus BC class
Only one Chem II class
Only one Physics C class
Repeat for basically every extra "advanced" elective
Not sure about Bio II i didn't take it
|
Well, he stated he wanted to be the one that knew the answer, not learn and understand it right away. Of course I'm not going to go and teach him something that his teachers are paid to teach him.
V=velocity (Vi = initial, Vf=final)
D=displacement
Do=original displacement from origin
t=time
a= acceleration
Notice that the 3rd equation doesn't need time.
To find the initial velocity, use the 3rd equation. The initial velocity is 0 in that equation, since you're starting from rest. The actual "initial velocity" they want is the "final velocity" in respect to this equation, since you're using it only to find the speed at which the ball leaves his hands. The acceleration is given, the distance is given (convert to meters). Solve for Vf.
To find the time, use the second equation. d=do+vit+1/2at^2. Do is 2.3-1.82 since that's where it was starting (above his head). d is given (1.82), you found the vi in the last problem, the acceleration is -9.8 m/s^2 because that is the acceleration of a body through gravity. So now you have everything to solve for time.
edit: Oops forgot 2nd one.
Equation is the 3rd one.
Vf is 0, since at the peak of the ball's trajectory, it's velocity is 0. The acceleration is still -9.8. The Vi, you already found in #1. So just solve for d.
Or you can use the energy way vAltyR described (the way I learned it was kinematics first, and energy second, but you can do either one).
|
whoaa @_@ confusing gimme a sec to write this all down
nice one vAltyR *posted above*
I'll compare answers when i'm done
Lmao, i spent like 10 minutes figuring out what the d value was, when i realized "OH, I'm Solving for D!" Doh >.<"
+ Show Spoiler [Solving for Vf] +Vf^2= Vi^2 + 2ad
Vf^2=(0)^2 + 2(45.7m/s^2)(.69m)
Vf^2= 31.553m^2/s^2
Vf=5.6154m/s
+ Show Spoiler [In process!] +0= 5.6154m/s+ (2)(-9,8m/s^2)(d) -d= -13.9846 eer D: meters? d= ??
+ Show Spoiler [inprocess] +-.48= -4.9t^2+5.6154t
eh eff it quadratic formula ftw gg
|
OK that's a better explanation but it's also like the most confusing possible way to do this problem...
|
United States24799 Posts
Taking Physics C without taking another physics class before is quite tough.
|
+ Show Spoiler +On September 24 2008 06:52 nevake wrote:
Well, he stated he wanted to be the one that knew the answer, not learn and understand it right away. Of course I'm not going to go and teach him something that his teachers are paid to teach him.
V=velocity (Vi = initial, Vf=final)
D=displacement
Do=original displacement from origin
t=time
a= acceleration
Notice that the 3rd equation doesn't need time.
To find the initial velocity, use the 3rd equation. The initial velocity is 0 in that equation, since you're starting from rest. The actual "initial velocity" they want is the "final velocity" in respect to this equation, since you're using it only to find the speed at which the ball leaves his hands. The acceleration is given, the distance is given (convert to meters). Solve for Vf.
To find the time, use the second equation. d=do+vit+1/2at^2. Do is 2.3-1.82 since that's where it was starting (above his head). d is given (1.82), you found the vi in the last problem, the acceleration is -9.8 m/s^2 because that is the acceleration of a body through gravity. So now you have everything to solve for time.
edit: Oops forgot 2nd one.
Equation is the 3rd one.
Vf is 0, since at the peak of the ball's trajectory, it's velocity is 0. The acceleration is still -9.8. The Vi, you already found in #1. So just solve for d.
Or you can use the energy way vAltyR described (the way I learned it was kinematics first, and energy second, but you can do either one).
I vaguely remember acceleration always being positive. However, What do i know Anyone else verify acceleration negative?
Ooh there's different ways to solve it?
|
On September 24 2008 07:17 micronesia wrote:
Taking Physics C without taking another physics class before is quite tough.
How'd you know i didn't take a physics class before this? D:
Mind reader, you are.
The units are driving me crazy... everything has to be perfect
and u can't use like SI units with... metrics and god damn how are you supposed to remember everything D::::
O well, i'm a glutton for pain ;D
|
It depends on which way you are saying is positive. Usually, I find that making UP positive is the easiest, therefore making acceleration negative (since when something falls, it's accelerating down). If you make acceleration positive, then you have to make Y distance negative since the acceleration is opposite to where the object is going (on the way up). It's much easier to picture that way. Yea taking physics C is hard then. Taking calculus and phys B before or during makes it much easier.
|
I'm taking Calc BC right now
|
d=do+vit+1/2at^2
1.82=2.3+5.6154*t+.5*-9.8*t^2
0= -4.9t^2+5.6154t+.48
|
United States24799 Posts
On September 24 2008 07:19 HeavOnEarth wrote:Show nested quote +On September 24 2008 07:17 micronesia wrote:
Taking Physics C without taking another physics class before is quite tough. How'd you know i didn't take a physics class before this? D: Mind reader, you are.
I can't tell if you are being sarcastic or not.
If yes: meh
If no: Kinda obvious from the OP.
|
The "D:" within lies many unsolved mysteries...
For me it's sarcasm
or just friendly joking around :O
|
Here's how I solved it (I think I did it all correctly), and I put each successive step in spoilers in case you wanna try it out yourself first:
You can ignore the weight, since gravity is always -9.8m/s^2 if down is negative.
So you have:
a=45.7m/s^2
g=-9.8m/s^2
initial v=0
distance to release=.69m converted from cm
To find the speed of the shot as it's released, use the third equation posted by nevake, and you'll get:
v^2= 0 + 2(45.7)(0.69)
+ Show Spoiler +v= 7.94 m/s Next I find the time it takes from the beginning to the releasing of the rock. To do that, I use the 1st equation, using the final velocity I found and the acceleration provided 7.94 - 9.8t time to release = + Show Spoiler +To find the time to the highest point, I also use the first equation. I know the final velocity is 0 because at the very top of the flight, the velocity is always zero, since to go from positive to negative velocity, it has to pass through 0 velocity. Also, the acceleration is -9.8m/s^2 since now that the man released the rock, only gravity is acting on it. So you get this equation: 0 = 7.94 - 9.8t time to top = + Show Spoiler +With the time from release to the highest point discovered, we can now use the 2nd equation to find the distance it takes to reach the top distance to top = 0.69 + 7.94(.81) -4.9(.81^2) = + Show Spoiler +To find the time it takes to reach the man's head, we again use the 2nd equation, since we have the distance, initial velocity, and acceleration: + Show Spoiler +
And there you go. This problem you will consider easy later on after you've practiced how to apply the three equations, and other ones will be pretty much the same, you just need to apply them differently as they add in angles and two directions.
edit: I'm solving these algebraically by the way, not with Calculus, since my school only has 3 students who took AP Calc taking AP Physics as seniors, so most other people don't know any Calculus. I looked at some of the other answers ppl posted and they're different than mine, so I dunno, I guess check with your teacher since I don't wanna spend more time on this XD. I did this the most straightforward way I could think of using the 3 equations, and since he was throwing a 16 pound rock at that acceleration for about half a bodylength, 3.91m to the top sounds about right compared to like 10m in one of the other problems.
o yea, and make sure to tell us what the real answer is lol
|
Will do. Probably update with answer tomorrow.
|
On September 24 2008 07:19 HeavOnEarth wrote:Show nested quote +On September 24 2008 07:17 micronesia wrote:
Taking Physics C without taking another physics class before is quite tough. How'd you know i didn't take a physics class before this? D: Mind reader, you are. The units are driving me crazy... everything has to be perfect and u can't use like SI units with... metrics and god damn how are you supposed to remember everything D:::: O well, i'm a glutton for pain ;D
He's a certified physics teacher
|
On September 24 2008 06:22 HeavOnEarth wrote:Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 45.7 m/s2 for a height 69.0 cm. He releases it at height 2.30 m above the ground. *Meaning the ball released his hand @ 2.30m above the ground* You may ignore air resistance. What is the speed of the shot when he releases it? How high above the ground does it go? How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.82 m above the ground? It's not really a homework problem; she just posted it on the board and said try it at home , see if you can solve it ; and i wanna be the cool kid that knows the answer Of course, I'm not blindly giving you this question without working on it, however i probably messed everything up :D I don't expect anyone to do it but... if you could point me in the general direction that would be pretty awesome [: . ;edit; Excuse my godawful drawing T_T ;edit2; Also 69cm = (69)1x10^-2 m or .69 m And wow TL is really helpful :DD Also, i suppose that when the ball is coming down ; g ( acceleration), due to gravity = ~9.81m/s^2 Ahh and also this wasn't a homework problem or anything she just gave it to us as a kind of eye opener , she's going to intro./ review it tomorrow in class. So far we've just been doing physics I review so we're starting kinematics, plus the hurricane set us back a week( no school) so that would explain the lack of movement in the class? D: Thanks for your help again !
please try to type the problem correctly so it can be solved.
you can't give something a constant acceleration by throwing it. constant acceleration means that it keeps accelerating, which isn't possible if you throw it.
if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely.
the problem should be straightforward if the 45.7 is velocity rather than acceleration. the object starts by having an upward velocity of 45.7. from then, each second that passes by, it loses 9.8 of its original velocity due to gravity. this means that it goes upwards for (45.7/9.8) seconds before it reaches its highest point.
=> so if while going upwards, its time spent is 45.7/9,8, and its average velocity is 45.7/2, you can figure out the distance.
=> now you can find the total height at the highest point. from there, you can find how long it takes for the object to reach a height of 1.82.
|
On September 24 2008 11:19 Polyphasic wrote:Show nested quote +On September 24 2008 06:22 HeavOnEarth wrote:Sam heaves a shot with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 45.7 m/s2 for a height 69.0 cm. He releases it at height 2.30 m above the ground. *Meaning the ball released his hand @ 2.30m above the ground* You may ignore air resistance. What is the speed of the shot when he releases it? How high above the ground does it go? How much time does he have to get out of its way before it returns to the height of the top of his head, a distance 1.82 m above the ground? It's not really a homework problem; she just posted it on the board and said try it at home , see if you can solve it ; and i wanna be the cool kid that knows the answer Of course, I'm not blindly giving you this question without working on it, however i probably messed everything up :D I don't expect anyone to do it but... if you could point me in the general direction that would be pretty awesome [: . ;edit; Excuse my godawful drawing T_T ;edit2; Also 69cm = (69)1x10^-2 m or .69 m And wow TL is really helpful :DD Also, i suppose that when the ball is coming down ; g ( acceleration), due to gravity = ~9.81m/s^2 Ahh and also this wasn't a homework problem or anything she just gave it to us as a kind of eye opener , she's going to intro./ review it tomorrow in class. So far we've just been doing physics I review so we're starting kinematics, plus the hurricane set us back a week( no school) so that would explain the lack of movement in the class? D: Thanks for your help again ! please try to type the problem correctly so it can be solved. you can't give something a constant acceleration by throwing it. constant acceleration means that it keeps accelerating, which isn't possible if you throw it. if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely. the problem should be straightforward if the 45.7 is velocity rather than acceleration. the object starts by having an upward velocity of 45.7. from then, each second that passes by, it loses 9.8 of its original velocity due to gravity. this means that it goes upwards for (45.7/9.8) seconds before it reaches its highest point. => so if while going upwards, its time spent is 45.7/9,8, and its average velocity is 45.7/2, you can figure out the distance. => now you can find the total height at the highest point. from there, you can find how long it takes for the object to reach a height of 1.82.
You say all this in light of a posted solution on the 1st page?
|
please try to type the problem correctly so it can be solved.
but but i typed out the 69cm , it only goes up to 69 cm
if it is as you say, that there is a constant upward acceleration of 45.7m/s2, (and if you take into account that downward acceleration due to gravity is 9.8m/s2), then the result is that the object will continue to accelerate upwards indefinitely.
noooh it stops after the .69m thing
im sry ill make it clearer even though i bolded it and drew a picture and everything
;edit; im soo confused now T.T
|
my bad. haha, in that case:
if the ball accelerates for .69 meters, after which its acceleration is 0 (presumably because it took .69 meters to throw the ball).
1- find the time it takes for the ball to go .69 meters if it starts at rest, and its acceleration is 45.7m/s2. (this is an equation you use)
2- after that, the scenario is that the ball starts from the end of the throw (starting point plus .69 meters) at the velocity you calculated earlier. from this point, the solution is same as my previous post.
=) don't take my toneless post too hard. i tend to be very friendly in real life.
|
Well i'll post all my answers and stuff + teachers solutions tomorrow! for now i finish my english work =/
|
United States24799 Posts
On September 24 2008 09:53 SayaSP wrote:Show nested quote +On September 24 2008 07:19 HeavOnEarth wrote:On September 24 2008 07:17 micronesia wrote:
Taking Physics C without taking another physics class before is quite tough. How'd you know i didn't take a physics class before this? D: Mind reader, you are. The units are driving me crazy... everything has to be perfect and u can't use like SI units with... metrics and god damn how are you supposed to remember everything D:::: O well, i'm a glutton for pain ;D He's a certified physics teacher
Only conditionally initially:
http://www.highered.nysed.gov/tcert/certificate/typesofcerts.htm#ci
|
|
|
|
|
|
EPT![[image loading]](http://i38.tinypic.com/16bcwi9.jpg)
