trouble using an equation in a function

Question

Write a program that determines how far and for how long a time a rock will travel when you throw it off a cliff. Click here to copy the file toss.txt to your desktop (right click the file name and choose Save as). The file contains the height of the cliff in meters.

The program will then:

1. Open the file toss.txt and read the cliff height into a double-precision variable, then echo print the value of the cliff height to the screen with an appropriate label.

2. Ask the user for the angle at which the rock is thrown (90 degrees is straight up, and 0 degrees is straight forward), and the velocity at which the rock is thrown (in miles per hour).

3. Check to make sure the angle is greater than or equal to 0 and less than or equal to 90. If it is not, the program terminates and prints an appropriate error message to the screen.

4. Check to make sure the velocity is less than or equal to 100 mph and greater than or equal to 0 mph. If it is not, the program terminates and prints an appropriate error message to the screen.

5. If the angle and velocity are valid, the program completes the calculations as follows:

   1. Converts miles per hour to meters per second.

   2. Converts the angle to radians.

   3. Calculates the time traveled using the following equations:

      where

   4. Calculates the distance traveled in the horizontal direction using:

6. Outputs the time and distance traveled in the horizontal direction to the screen with appropriate labels.

7. Prints an appropriate message telling the user if the distance traveled in the horizontal direction was greater than, less than, or equal to the height of the cliff.

    /* This program */ using namespace std; #include<iostream> #include<cmath> #include<iomanip> #include<fstream> int readit (); int calcit (double, double, double); int main() { readit (); system ("pause"); return 0; } int readit () { double hite, angl, v; ifstream datain ( "toss.txt" ); datain >> hite; cout << "The cliff height is " << hite << " meters"<< endl; cout << "Enter the angle in degrees (from horizontal) the rock is thrown: " << endl; cin >> angl; if (angl>=0 && angl<=90) { cout << endl << "The angle you have entered is "<<angl<< endl <<endl; } else { cout << "The angle you have entered is not acceptable" << endl; return 0; } cout << "Enter the velocity in mph the rock is thrown: " << endl; cin >> v; if (v>=0 && v<=100) { cout << endl << "The velocity at which the rock is thrown is "<<v<< " mph" << endl << endl; } else { cout << "The velocity you have entered is not acceptable" << endl; return 0; } calcit (hite, angl, v); } int calcit (double hite, double angl, double v) { double tyme, dist; v = v * (1609.344/3600); angl = angl*(M_PI/180); tyme = -v*sin(angl) + (sqrt((v*sin(angl)*v*sin(angl)) + 2*9.8*hite)/9.8) + (2*(v*sin(angl))/9.8); dist = (tyme * v) * cos(angl); cout << tyme << " " << dist <<endl; } 

I am trying to get the correct time the rock is traveling before it hits the ground but i keep getting incorrect answers. I am not sure if i am turning the equation to figure out the time the rock will be in the air until impact into c++ language right. any have any ideas??? i really need to finish this damn project.

Answer 1

Starting from the equation for the y (height above 0) for the rock we have

y = h + v*sin(a)*t - g/2*t^2

which transforms into

g/2 T^2 - v*sin(a)*T - h == 0

when we solve for the final condition y(T)=0 .

This yields

T = v*sin(a)/g + sqrt(v*sin(a)*v*sin(a) + 2*g*h)/g

I just can't figure out where the first part -v*sin(angl) in your equation comes from. Everything else looks just fine. So it seems not to be with your code but with the equation you started.

Answer 2

I would suggest a few things to "clean up" the code a bit:

• If functions return int ensure that they do really return something. (main doesn't have to but other functions do).

• Calculate v * sin(ang1) once then use it in your formula thereafter. Not only more efficient but will make your code clearer.

• Like you have given Pi a "constant", do that with other numbers you are using like 9.8 (gravitational force?)

Answer 3

The equation you want is:

s =ut + 1/2 at^2

s = Total distance traveled.   (Height of the cliff)
u = Starting velocity          (In your case negative as you are throwing
                                away from the target. And take into account
                                that not all the starting velocity is away
                                from the target (eg angle 0 mean u = 0))
a = acceleration               (9.81 m/s2)
t = time                       (The value you want to calculate).

Rearrange the formula to solve for t

To find the solution for t where s = 0...
This formula is you basic quadratic:

y = a.x^2 + b.x + c

Where:
    x/y     are variables.
    a/b/c   are constants.

The solution for a quadratic equation where y is 0 is:

x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

Notice the ± symbol. There are actually two solutions to the problem.
You should be able to deduce which one is correct for you as the other
is probably negative.

In your particular case the map is:

   x ==> t
   y ==> 0

   a ==> 1/2.g
   b ==> u
   c ==> -s

Answer 4

If you have a confusing formula in the code, just introduce more variable names until the meaning becomes obvious. So long as you don't reassign different values to the same variables, this will not make the program confusing.

int calcit (double hite_meters, double angl_deg, double v_mph)
{
    double const gravity = 9.8;
    double v_ms = v_mph * (1609.344/3600);
    double angl_rad = angl_deg * (M_PI/180);
    double v_vertical = v_ms * sin( angl_rad );
    double time_up = v_vertical / gravity; // [m/s] / [m/s^2] = [s]
    double time_down_over_cliff = time_up;
    // use quadratic formula t = ( -v - ( v^2 - 4gd )^1/2 ) / 2g:
    double time_under_cliff = ( - v_vertical
         - sqrt( ( v_vertical * v_vertical )
               - ( 4 * - gravity * hite_meters ) ) // negative gravity = down
        ) / ( 2 * - gravity ); // ( [m/s] + ([m/s]^2 - [m/s^2]*[m])^1/2 ) / [m/s^2]
                             // = [m/s] / [m/s^2] = [s]
    double time_total = time_up + time_down_over_cliff + time_under_cliff;
    double v_horizontal = v_ms * cos( angl_rad );
    double dist_horizontal = v_ms * time_total;

    cout << time_total << " " << dist_horizontal <<endl;
}

Every line of code produces a new, relevant piece of information. When converting to a new unit, I introduce a new variable with a new name. Formulas involving more than one unit get the unit types explained in a comment. This should help turn up unit conversion errors which otherwise I can't help you catch.

Writing this kind of code involves more typing, but the time saved on head-scratching and asking for help more than makes up for it.

The program itself is not any less efficient. More importantly, it may be easily modified, so it won't turn into an inefficient mess after a few revisions.