### create a function file that calculates the trajectory of a projectile. The inputs to the function are the initial velocity and the angle at which the projectile is fired. The outputs from the function are the maximum height and distance. In addition, the function generates a plot of the trajectory. Use the function to calculate the trajectory of a projectile that is fired at a velocity of 230 m/sec at an angle of 39deg.

Problem:
create a function file that calculates the trajectory of a projectile. The inputs to the function are the initial velocity and the angle at which the projectile is fired. The outputs from the function are the maximum height and distance. In addition, the function generates a plot of the trajectory. Use the function to calculate the trajectory of a projectile that is fired at a velocity of 230 m/sec at an angle of 39deg.

USING MATLAB SOFTWARE Version 7.0

SOLUTION:-

%Write it in script file
function[hmax,dmax]=trajectory(V0,theta)
g=9.81;
V0x=V0*cos(theta);
V0y=V0*sin(theta);
thmax=V0y/g;
hmax=V0y^2/(2*g);
ttotal=2*thmax
dmax=V0x*ttotal
% creating a trajectory plot
tplot=linspace(0,ttotal,200);
x=V0x*tplot;
y=V0y*tplot-0.5*g*tplot.^2;
plot(x,y)
xlabel('distance (mt)')
ylabel('height(mt)')
title('projectile of Rocket')

>> [height distance]=trajectory(230,39)        %In command window

ttotal =

45.1933

dmax =

2.7716e+003

height =

2.5045e+003

distance =

2.7716e+003 Trajectory of profile

### The co-efficient of friction mu can be determined in an experiment by measuring the force F required to move a mass m. when F is measured and m is known the coefficient of friction can be calculated by mu=F/mg (g=9.81 m/sec2) results from measuring F in six tests are given in the table below. Determine the coefficient of friction in each test, and average from all tests.

Problem:
The co-efficient of friction mu can be determined in an experiment by measuring the force F required to move a mass m. when F is measured and m is known the coefficient of friction can be calculated by mu=F/mg (g=9.81 m/sec2) results from measuring F in six tests are given in the table below. Determine the coefficient of friction in each test, and average from all tests.

 Test # 1 2 3 4 5 6 Mass m-kg 2 4 5 10 20 50 Force F -N 12.5 23.5 30 61 117 294

USING MATLAB SOFTWARE

SOLUTION:

%IN COMMAND WINDOW

>> m=[2 3 5 10 20 50];
>>  f=[12.5 23.5 50 61 117 294];
>>  mu=f./(m*9.8)

mu =

0.6378    0.7993    1.0204    0.6224    0.5969    0.6000

>> mu_avg=mean(mu)

mu_avg =

0.7128

### Three forces are applied to a bracket as shown. Determine the total (equivalent) force applied to the barcket

Problem:

% Three forces are applied to a bracket as shown. Determine the total (equivalent) force applied to the barcket%

%IN COMMAND WINDOW

>> F1m=400;F2m=500;F3m=700;
>> Th1=-20*pi/180;Th2=30*pi/180;Th3=143*pi/180;
>> F1=F1m*[cos(Th1) sin(Th1)]

F1 =

375.8770 -136.8081

>> F2=F2m*[cos(Th2) sin(Th2)]

F2 =

433.0127  250.0000

>> F3=F3m*[cos(Th3) sin(Th3)]

F3 =

-559.0449  421.2705

>> Ftotal=F1+F2+F3

Ftotal =

249.8449  534.4625

>> Ftotalm=sqrt(Ftotal(1)^2+Ftotal(2)^2)

Ftotalm =

589.9768

>> Th=(180/pi)*tan(Ftotal(2)/Ftotal(1))

Th =

-89.7088

### To determine Nearest of temp after 3 hours

Problem: To determine Nearest of temp after 3 hours..

>>% To determine Nearest of temp after 3 hours..%
>> Ts=38;T0=120;K=0.45;t=3;
>> T=round(Ts+(T0-Ts)*exp(-K*t))

T =

59

### A trigonometric identity is given by cos2(x/2)=(tanx+sinx)/2.tanx verify that the identity is correct by calculating each side of the equation substituting x=pi/5

Problem :
A trigonometric identity is given by cos2(x/2)=(tanx+sinx)/2.tanx verify that the identity is correct by calculating each side of the equation substituting x=pi/5

>> %A trigonometric identity is given by cos2(x/2)=(tanx+sinx)/2.tanx verify that the identity is correct by calculating each side of the equation substituting x=pi/5%
>> x=pi/5;
>> LHS=cos(x/2)^2

LHS =

0.9045

>> RHS=(tan(x)+sin(x))/(2*tan(x))

RHS =

0.9045

### c program to multiplication of two 3*3 matrices

/* c program to multiplication of two 3*3 matrices*/

#include<stdio.h>
#include<conio.h>
void main()
{
int m1,m2,prod;
int i,j,k;
clrscr();
printf("enter the first matrix : \n");
for(i=0;i<3;i++)
for(j=0;j<3;j++)
scanf("%d",&m1[i][j]);
printf("enter 2nd matrix : \n");
for(i=0;i<3;i++)
for(j=0;j<3;j++)
scanf("%d",&m2[i][j]);

/* multiplication of matrix */
for(i=0;i<3;i++)
for(j=0;j<3;j++)

{
prod[i][j]=0;
for(k=0;k<3;k++)
prod[i][j]=prod[i][j]+m2[i][k]*m2[k][j];
}

printf("\n\n product matrix\n");
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%d\t",prod[i][j]);
printf("\n");
}
getch();
}

enter the first matrix
4     4     4
4     4     4
4     4     4
enter 2nd matrix
4     4     4
4     4     4
4     4     4
product matrix
48    48    48
48    48    48
48    48    48

### c program to subtract two 3*3 matrices

/*c program to subtract two 3*3 matrices*/
#include<stdio.h>
#include<conio.h>
void main()
{
int m1,m2,msub;
int i,j;
clrscr();
printf("type in 9 integer number of first matrix:\n");
for(i=0;i<3;++i)
for(j=0;j<3;++j)
scanf("%d",&m1[i][j]);
printf("\n type in 9 integer number of first matrix:\n");
for(i=0;i<3;++i)
for(j=0;j<3;++j)
scanf("%d",&m2[i][j]);

/*matrix subtraction proceeds as follows...*/
for(i=0;i<3;++i)
for(j=0;j<3;++j)
msub[i][j]=m1[i][j]-m2[i][j];
/*subtraction matrix as follows...*/
for(i=0;i<3;++i)
{
for(j=0;j<3;++j)
printf("%3d",msub[i][j]);
printf("\n");
}
getch();
}

type in 9 integer number first matrix:
1 2 3
3 2 1
4 5 6

type in 9 integer number of second matrix
6 5 4
3 2 1
2 3 1

Subscription of matrix
-5 -3 -1
0  0  0
2  2  5

### c program to add two 3*3 matrices

/*c program to add two 3*3 matrices*/
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j;
clrscr();
printf("type in 9 integer number of first matrix:\n");
for(i=0;i<3;++i)
for(j=0;j<3;++j)
scanf("%d",&m1[i][j]);
printf("\n type in 9 integer number of first matrix:\n");
for(i=0;i<3;++i)
for(j=0;j<3;++j)
scanf("%d",&m2[i][j]);

for(i=0;i<3;++i)
for(j=0;j<3;++j)
for(i=0;i<3;++i)
{
for(j=0;j<3;++j)
printf("\n");
}
getch();
}

type in 9 number of first matrix:
1 2 3
1 2 4
2 1 3

type in 9 integer number of first matrix:
8 6 5
7 8 9
4 5 6

9  8  8
8 10 13
6  6  9

### to arrange number descending using bubble sort technique

/*to arrange number descending using bubble sort technique*/
#include<stdio.h>
#include<conio.h>
void main()
{
int num;
int loop,i;
int last=10;
int exchange;
int temp;
clrscr();

printf("enter the 10 number\n");
for(i=0;i<10;++i)
scanf("%d",&num[i]);
for(loop=0;loop<9;++loop)
{
exchange=0;
for(i=0;i<(last-1);i++)
if(num[i]>num[i+1])
{
temp=num[i];
num[i]=num[i+1];
num[i+1]=temp;
exchange=exchange+1;
}
if(exchange==0)
break;
else
last=(last-1);
}
printf("number is in ascending order: \n");
for(i=0;i<10;i++)
printf("%d\n",num[i]);
getch();
}

Enter the 10 number
88
0
14
1
6
9
6
94
52
22
Number is in descending order:
94
88
52
22
14
9
6
6
1
0

### to arrange number ascending using bubble sort techniqu

/*to arrange number ascending using bubble sort technique*/
#include<stdio.h>
#include<conio.h>
void main()
{
int num;
int loop,i;
int last=10;
int exchange;
int temp;
clrscr();
printf("enter the 10 number\n");
for(i=0;i<10;++i)
scanf("%d",&num[i]);
for(loop=0;loop<9;++loop)
{
exchange=0;
for(i=0;i<(last-1);i++)
if(num[i]<num[i+1])
{
temp=num[i];
num[i]=num[i+1];
num[i+1]=temp;
exchange=exchange+1;
}
if(exchange==0)
break;
else
last=(last-1);
}
printf("number is in descending order: \n");
for(i=0;i<10;i++)
printf("%d\n",num[i]);
getch();
}

Enter  the 10 number
25
20
6
14
12
32
4
6
8
9
Number is in ascending order:
4
6
6
8
9
12
14
20
25
32

### sum and average of 3 numbers

/*sum and average of 3 numbers*/
#include <stdio.h>
#include <conio.h>
void main()
{
int n1,n2,n3;
float sum,average;
printf("enter 3 integer numbers\n");
sum=(n1+n2+n3);
average=sum/3;
printf("The sum of 3 numbers is=%f\n,sum);
printf=("The average of numbers=%f/n",average);
getch()

}

out put
12
25
33
sum of 3 numbers=70.000000
avg of 3 numbers=23.333334

### Largest of 3 number

/*Largest of 3 number */
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b,c;
printf("enter the 3 numbers\n");
scanf("%d %d %d",&a,&b,&c);
if(a>b && a>c)
printf("a is greater\n");
else
if(b>a && b>c)
printf("b is greater");
else
printf("c is greater");

getch();

}

out put
enter the 3 numbers
9
7
20
c is greater

### To find whether a number is prime or not

#include<stdio.h>
#include<conio.h>
main()
{
int i,number,remainder,interval;
clrscr();
/*program to test the prime property of a given number*/
printf("type in the number to be tested\n");
scanf("%d",&number);
interval=number/2;
for(i=2; i<=interval; ++i)
{
remainder=number%i;
if(remainder==0)
break;
}
if(remainder!=0)
printf("given number %d is a PRIME NUMBER\n",number);
else
printf("given number %d is not a PRIME NUMBER\n",number);
}

OUTPUT

type in the number to be tested
43
given number 43 is a PRIME NUMBER

### finding the roots of the quadratic equation using switch-statement

#include<stdio.h>
#include<conio.h>
void main()
{
int option;
float a,b,c,d;
float root1,root2,r1,r2;
clrscr();
printf("enterthe value of a,b,c\n");
scanf("%f%f%f",&a,&b,&c);
if (a==0||b==0||c==0)
{
printf("\n error_we cannot analyse");
exit();
}
d=(b*b-4*a*c);

if(d>0)
option=1;
else

if(d<0)
option=2;
else
option=3;
switch(option)
{
case1:
printf("roots are real & distinct\n");
root1=(-b+sqrt(d))/2*a;
root2=(-b-sqrt(d))/2*a;
printf("\n root1=%f root2=%f\n",root1,root2);
break;

case 2:
printf("\n roots are imaginary & complex\n");
r1=-b/2*a;
r2=sqrt(abs(d))/2*a;
printf("root1=%f+%f\n",r1,r2);

case 3:
printf("\n roots are real & equal\n");
root1=root2=-b/2*a;
printf("root1=%f\n,root2=%f\n",root1,root2);
break;
}
getch();
}

OUTPUT

enter the value of a,b,c
22,33,44
roots are imaginary & complex
root1=-363.000000+27962.000000

roots are real & equal
root2=-363.000000

### whether number is even or odd

/*whether number is even or odd*/
#include<stdio.h>
#include<conio.h>
void main()
{
int num,remainder;
clrscr();
printf("enter the number\n");
scanf("%d",&num);
remainder=num%2;
if (remainder==0)
printf("the number is even\n");
else
printf("the number is odd\n");
getch();
}

OUTPUT

enter the number
121
the number is odd

### sum of digit of given number

/*sum of digit of given number*/
#include<stdio.h>
#include<conio.h>
void main()
{
int number,sum=0,r;
clrscr();
printf("enter the number\n");
scanf("%d",&number);
do
{
r=number%10;
sum=sum+r;
number=number/10;
}
while(number!=0);
printf("sum of digit is=%d\n",sum);
getch();
}

OUTPUT
enter the number
123
sum of digit is=6