- #1

polka129

- 43

- 0

hello people...i have beeen given this projectile problem to be solved numerically in FORTRAN..i have coded it using runge0kutta 4th order ...now the thing is that i have not been given the end points of time,,;ie. the range.. and i am asked to

a)the maximum height attained by the projectile

what is to b added to the existing code to achive this?

b)the time required to reach the maximum height

what is to b added to the existing code to achive this?

c)the time required to return to the original elevation..

what is to b added to the existing code to achive this?

i did fortran 4 years ago...havent beeen in practice..this was the lone effort i cud recall...please help...

i have attached the question and the code..

heres the code

CODE

!**********************************************************************

! *

! projectile question *

! *

!***********************************************************************

DATA M,G,C/10.0,9.80665,0.1/

!FIRST i HAVE REDUCED THE GIVEN EQUATION IN TWO LINEAR 1ST-ORDER EQUATIONS

!FOR THE SOLUTION TO PROCEED

F1(T,U1,Y2) = U1

F2(T,U1,Y2) = (-M*G-C*U1**2)/M

!ASSIGNMENT OF VALUES TO CONSTANTS

WRITE(*,*) 'Input left and right endpoints separated by'

WRITE(*,*) 'blank'

WRITE(*,*) ' '

READ(*,*) A, B

WRITE(*,*) 'Input the two initial conditions.'

WRITE(*,*) ' '

READ(*,*) ALPHA1, ALPHA2

WRITE(6,*) 'Input a positive integer for the number'

WRITE(6,*) 'of subintervals '

WRITE(6,*) ' '

READ(5,*) N

WRITE(*,6)

6 FORMAT(12X,'t(i)',11X,'w1(i)',11X,'w2(i)')

H=(B-A)/N

T=A

! the initiaal conditions

W1=ALPHA1

W2=ALPHA2

WRITE(*,1) T,W1,W2

!RK PARAMETER EVALUATIONS HERE

DO 110 I=1,N

X11=H*F1(T,W1,W2)

X12=H*F2(T,W1,W2)

X21=H*F1(T+H/2,W1+X11/2,W2+X12/2)

X22=H*F2(T+H/2,W1+X11/2,W2+X12/2)

X31=H*F1(T+H/2,W1+X21/2,W2+X22/2)

X32=H*F2(T+H/2,W1+X21/2,W2+X22/2)

X41=H*F1(T+H,W1+X31,W2+X32)

X42=H*F2(T+H,W1+X31,W2+X32)

W1=W1+(X11+2*X21+2*X31+X41)/6

W2=W2+(X12+2*X22+2*X32+X42)/6

T=A+I*H

WRITE(*,1) T,W1,W2

110 CONTINUE

STOP

1 FORMAT(3(1X,E15.8))

END

a)the maximum height attained by the projectile

what is to b added to the existing code to achive this?

b)the time required to reach the maximum height

what is to b added to the existing code to achive this?

c)the time required to return to the original elevation..

what is to b added to the existing code to achive this?

i did fortran 4 years ago...havent beeen in practice..this was the lone effort i cud recall...please help...

i have attached the question and the code..

heres the code

CODE

!**********************************************************************

! *

! projectile question *

! *

!***********************************************************************

DATA M,G,C/10.0,9.80665,0.1/

!FIRST i HAVE REDUCED THE GIVEN EQUATION IN TWO LINEAR 1ST-ORDER EQUATIONS

!FOR THE SOLUTION TO PROCEED

F1(T,U1,Y2) = U1

F2(T,U1,Y2) = (-M*G-C*U1**2)/M

!ASSIGNMENT OF VALUES TO CONSTANTS

WRITE(*,*) 'Input left and right endpoints separated by'

WRITE(*,*) 'blank'

WRITE(*,*) ' '

READ(*,*) A, B

WRITE(*,*) 'Input the two initial conditions.'

WRITE(*,*) ' '

READ(*,*) ALPHA1, ALPHA2

WRITE(6,*) 'Input a positive integer for the number'

WRITE(6,*) 'of subintervals '

WRITE(6,*) ' '

READ(5,*) N

WRITE(*,6)

6 FORMAT(12X,'t(i)',11X,'w1(i)',11X,'w2(i)')

H=(B-A)/N

T=A

! the initiaal conditions

W1=ALPHA1

W2=ALPHA2

WRITE(*,1) T,W1,W2

!RK PARAMETER EVALUATIONS HERE

DO 110 I=1,N

X11=H*F1(T,W1,W2)

X12=H*F2(T,W1,W2)

X21=H*F1(T+H/2,W1+X11/2,W2+X12/2)

X22=H*F2(T+H/2,W1+X11/2,W2+X12/2)

X31=H*F1(T+H/2,W1+X21/2,W2+X22/2)

X32=H*F2(T+H/2,W1+X21/2,W2+X22/2)

X41=H*F1(T+H,W1+X31,W2+X32)

X42=H*F2(T+H,W1+X31,W2+X32)

W1=W1+(X11+2*X21+2*X31+X41)/6

W2=W2+(X12+2*X22+2*X32+X42)/6

T=A+I*H

WRITE(*,1) T,W1,W2

110 CONTINUE

STOP

1 FORMAT(3(1X,E15.8))

END