LiveWire Network Peer Answers Peer Support Teen Forums Tech Forums College Forums 549 users online 211558 members 1554 active today Advertise Here Sign In
TeenCollegeTechPhotos | Quizzes | LiveSecret | Memberlist | Dictionary | News | FAQ
Member Spotlight
purple smoke
Cool Things: sex. anddddd.....sex
Mood: Flirty
You have 1 new message.
Emergency Help
Until you sign up you can't do much. Yes, it's free.

Sign Up Now
Membername:
Password:
Already have an account?
Invite Friends
Active Members
Groups
Contests
Moderators
6 online / 29 MPM
Fresh Topics
  LiveWire / Teen Forums / School & Homework Connection / Viewing Topic

Physics Problem
Replies: 9Last Post Jan. 4 5:13pm by nullPointerException
Welcome to LiveWire!
We're Stronger Together.
Join the Community
Single page for this topic Email Print Favorite
( morgan82 )


Dairy Product Addict
Reply
An object is launched with an initial velocity of 46 m/s at and angle of 57.0 degrees.

What is the max height it will reach?

Can someone help me out?


4:49 pm on Jan. 4, 2009 | Joined: April 2007 | Days Active: 346
Join to learn more about morgan82 United States | Posts: 1,218 | Points: 4,687
Frubeling

Guru

Patron
Reply
SHIT, Ive got an exam on this in just over a week, I CANT DO IT YET

-------
http://www.last.fm/user/Frubeling

4:51 pm on Jan. 4, 2009 | Joined: June 2008 | Days Active: 322
Join to learn more about Frubeling England, United Kingdom | Posts: 21,341 | Points: 21,790
snowfish


Swami

Patron
Reply
if you have a graphing calculator, you can just graph it with and find the peak of the parabola. Otherwise I think you break it down into vertical and horizontal components and solve it like a vector.

I'm sorry I haven't taken a physics class in two years.

-------
LW's resident eccentric radical.


4:51 pm on Jan. 4, 2009 | Joined: Feb. 2006 | Days Active: 673
Join to learn more about snowfish Massachusetts, United States | Female | Posts: 11,650 | Points: 25,750
Fearfulteen


Executive
Reply
i'm only in 8th grade!!!!!

-------
i'm proud to be a redneck!

4:52 pm on Jan. 4, 2009 | Joined: Dec. 2008 | Days Active: 137
Join to learn more about Fearfulteen Virginia, United States | Straight Female | Posts: 2,424 | Points: 3,872
commentary


Dairy Product Addict
Reply
I know it has to do with the sin of the 57 degree angle..

4:53 pm on Jan. 4, 2009 | Joined: Nov. 2008 | Days Active: 33
Join to learn more about commentary United States | Posts: 150 | Points: 1,130
bass09


Advisor
Reply
ya   about that

-------
What A Life

4:55 pm on Jan. 4, 2009 | Joined: Jan. 2009 | Days Active: 12
Join to learn more about bass09 Minnesota, United States | Straight Male | Posts: 302 | Points: 441
Anonymous

Reply
hopefully this is right.  

You gotta find the y component of the velocity, so you would do sin(57) x 46 m/s.  Then you get the y component speed.  Then you need to find the total kinetic energy the object has, using the equation 1/2(mass)(velocity)^2.  You don't have mass, but it will cancel out later anyway, so its ok.

So then you have the kinetic energy which is   "blahblahblah(mass)"  and you set that equal to (mass)(accel due to gravity)(height).  The kinetic energy must equal the potential energy at the highest point.

So the masses from each side of the equation cancel out.

Then you get some number = (accel due to gravity)(height).

So then you divide the left by the accel due to gravity, and you get the height

sorry i don't have a calculator so i can't do the sin 57 part.


4:57 pm on Jan. 4, 2009
blufindr


Asynchronous duality

Patron
Reply
From the ground, right?

sin(57) = y/46
y=46*sin(57)
y=38.58ms^-1

You know the velocity at the top is going to be 0ms^-1 and acceleration is -9.8ms^-2
So
v^2=u^2+2ax
0=148.83-19.6x
x=148.83/19.6
x=75.94m

So the object goes up 75.94 metres.

-------
Two hearts, beating as one;
Keeping time
Until they are no longer
Apart


5:02 pm on Jan. 4, 2009 | Joined: April 2008 | Days Active: 316
Join to learn more about blufindr Australia | Label Free Female | Posts: 36,248 | Points: 41,967
DTXBrian

Technician
Reply
To be able to determine the maximum height of the object, you have to know how fast the initial vertical velocity is.  To do that, you take the sine of 57 degrees, which is .8387 and multiply it by the total initial velocity, and you get that the object is travelling approximately 38.6 m/s vertically.

Now, you go back to your standard kinematic equations...

The maximum height will be reached at the exact point the object is no longer traveling upwards, so using the equation V=Vo + AT, you can solve for T (time)

0=38.6 + -9.8T

9.8T = 38.6

9.8T/9.8 = 38.6/9.8

T = 3.94 seconds

Now, go back to the equations...

S = So + VoT + 1/2AT^2

S = 0 + 38.6x3.94 + .5x-9.8x3.94^2

S = 152.8 + -4.9x15.52

S = 152.8 - 76.05

S = 76.75 meters in the air.

If you need more clarification, message me.


5:08 pm on Jan. 4, 2009 | Joined: Nov. 2008 | Days Active: 33
Join to learn more about DTXBrian United States | Posts: 1 | Points: 331
nullPointerException


Dairy Product Addict
Reply
y component of velocity = vsin(angle from horizontal)  
Y compnentof velocity = (46m/s)(sin(57))
(V final)^2 = (V initial)^2 +2ad
((V final)^2 - (inital v of y component)^2)/2a = d
((0 m/s)^2 - (46m/s)(sin(57))^2) / (-2g) = d
((0 m/s)^2 - (46m/s)(sin(57))^2) / (-2 (9.81 m/s^2) )= d


solve, dont have a calculator on hand


5:13 pm on Jan. 4, 2009 | Joined: May 2008 | Days Active: 150
Join to learn more about nullPointerException New York, United States | Straight Male | Posts: 976 | Points: 2,537
Single page for this topic Email Print Favorite

Quick Reply

You are signed in as our guest.

Looking for something else?
 

  LiveWire / Teen Forums / School & Homework Connection / Viewing Topic