physics!!

[nid404]

This is crazy....I've tried everything to come to the answer!!

Please help

check the attachment for the diagram

At the moment of release, the shot is 2.3m above the ground. The shot has a mass of 7.3kg. Calculate the kinetic energy of the shot at the moment of release....
How on earth is the ans 1713J  am I using the wrong velocity...we're supposed to use the resultant right? in the direction of 12m/s

[nid404]

Help? Pleeaasseeeee

[nid404]

http://books.google.co.in/books?id=4GUxeWap9qcC&pg=PA110&lpg=PA110&dq=At+the+moment+of+release,+the+shot+is+2.3m+above+the+ground.+The+shot+has+a+mass+of+7.3kg.+Calculate+the+kinetic+energy+of+the+shot+at+the+moment+of+release&source=bl&ots=BiPH-ngkim&sig=4FJ2IoVev9CwHiz_DWyXtG_f6WQ&hl=en&ei=9crGS7qlIpK4rAevs53zCQ&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAYQ6AEwAA#v=onepage&q=At%20the%20moment%20of%20release%2C%20the%20shot%20is%202.3m%20above%20the%20ground.%20The%20shot%20has%20a%20mass%20of%207.3kg.%20Calculate%20the%20kinetic%20energy%20of%20the%20shot%20at%20the%20moment%20of%20release&f=false

here's the complete question
q 52

[T.Q]

This is crazy....I've tried everything to come to the answer!!

Please help

check the attachment for the diagram

At the moment of release, the shot is 2.3m above the ground. The shot has a mass of 7.3kg. Calculate the kinetic energy of the shot at the moment of release....
How on earth is the ans 1713J  am I using the wrong velocity...we're supposed to use the resultant right? in the direction of 12m/s

the right answer is 713J not 1713J

because K.E=0.5(7.3)(13.97....)^2
            K.E=712.919....... J
           

[nid404]

I got the same answer....

But the back of the book says 1713 

[T.Q]

I got the same answer....

But the back of the book says 1713 

it might be a printing error

add the 1  to 713 by  mistake 

[nid404]

Let's hope so...


Makes me feel good   

Thanks tho

[student]

How did you get 13.97 anyway?
12+3cos54=13.76..

[T.Q]

How did you get 13.97 anyway?
12+3cos54=13.76..

the resultant velocity


v=(12cos54+3)^2 + (12sin54)^2

then take the sqr root of the ans

[Amr Fouad]

dont worry nid..it's 713 definitely..

[student]

Oh.. i see.
But if we resolve 3 in the direction of 12 (12+3cos54).. is that a slight variation of answer or a wrong way?

[Amr Fouad]

a wrong way.. 3 is already in a horizontal direction..dont confuse your self

[T.Q]

Oh.. i see.
But if we resolve 3 in the direction of 12 (12+3cos54).. is that a slight variation of answer or a wrong way?

u cant resolve 3  , because  3  is in the horizontal direction  

but the 12 , can be resolved in the vertical and horizontal directions  

[student]

If I take the direction of 12 as y-axis?
And add the component of 3 in that direction?

[T.Q]

If I take the direction of 12 as y-axis?
And add the component of 3 in that direction?

u cant , the direction of the y-axis is vertical

12 is not vertical