Final velocity need help

  • #1
final velocity...need help!!

Homework Statement



An object is to be thrown through a window 3.25 m high a distance 8 m away. It is launched at an angle 75 degree above the horizontal.

The initial launch speed at the specified angle which will allow the projectile to pass through the window is 13.3 m/s.

What is the final speed of the object (that is, the speed at which it flies through the window)?
Y= height
Yo= initial height
Voy= initial speed
Vf= final velocity
g= gravity
t= time

Homework Equations


1)Y= Yo+Voy*t-(1/2)gt^2
2)Vf= Vo+at


3. The Attempt at a Solution

By using the y= yo+Voy*t-(1/2)gt^2 formula,
= 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2
I found the time t=2.85 s

But then the answer i got by using the formula Vf= Vo+at
= 13.3+ (-9.81)(2.85)
= -14.7 m/s
IS wrong!!What is my mistake??
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi theoronhalde ! Welcome to PF! :smile:

(try using the X2 and X2 tags just above the Reply box :wink:)
An object is to be thrown through a window 3.25 m high a distance 8 m away. It is launched at an angle 75 degree above the horizontal.

The initial launch speed at the specified angle which will allow the projectile to pass through the window is 13.3 m/s.

What is the final speed of the object (that is, the speed at which it flies through the window)?

But then the answer i got by using the formula Vf = Vo+at
= 13.3+ (-9.81)(2.85)
= -14.7 m/s
IS wrong!!What is my mistake??

Vfy = Voy + at :wink:
 
  • #3


hi!! thx a lot!!

Not sure i understand your answer? Do i have to use the formula you wrote instead of the one i was using to fnd the final velocty? I'm a bit confuse...
 
Last edited:
  • #4
tiny-tim
Science Advisor
Homework Helper
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Yes … you have to deal with the y-components separately. :smile:

(the x-component of velocity, of course is constant :wink:)
 
  • #5


ok s it's going to be:

Vfy= Voy+at
= 13.3sin75+ (-9.81)(2.85)
= -15.1 m/s

But how come the horizontal velocity be constant? And am I suppose to use
V=Sqrt((Vox^2)+(Vfy^2)) to find that final speed?
 
  • #6
tiny-tim
Science Advisor
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The horizontal acceleration is zero, so the horizontal velocity must be constant.

And yes, use Pythagoras. :smile:
 
  • #7


oh ok....i undestand what you are saying but i still end up with a wrong answer...here is what i did:

Vfx= Vox + at (a=0)
=13.3cos75

Vfy= Voy+at
= 13.3sin75 + (9.81)(2.85)
= -15.1


Vf= SQRT( Vfy^2 + Vfx^2)
= 15.5 m/s

The applet says that the answer is wrong :S ??
 
  • #8
tiny-tim
Science Advisor
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I haven't checked how you got t using Vy, but when I do it using Vx, I don't get 2.85.
 
  • #9


oh...i'm so lost once again. I tought we should get the time By using:
y= yo+Voy*t-(1/2)gt^2 formula,
= 3.25+ (13.3sin75)t-(1/2)(-9.81)t^2
If not can you explain me how to find the time and how to get the components Vfy and Vfx.
 

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