|
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 States24761 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.
|
|
|
|
|
|
EPT![[image loading]](http://i38.tinypic.com/16bcwi9.jpg)
