How to compute the position of an object after free falling for 10 seconds [closed]

Question

I'm given this basic code:

public class GravityCalculator {
    public static void main(String[] args) {
        double gravity = -9.81; //Earth's gravity in m/s^2
        double initialVelocity = 0.0;
        double fallingTime = 10.0;
        double initialPosition = 0.0;
        double finalPosition = 0.0;
        System.out.println("The object's position after " + fallingTime + " seconds is " + finalPosition+ "m");
        // the output is The object's position after 10.0 seconds is 0.0m
    }
}

And I'm told to modify the program to compute the position of an object falling for 10 seconds, using this formula:

x(t) = 0.5 * at^2 + v(t) + x
a = acceleration = -9.81 m/s
t = time (in seconds) = 10
v = initial velocity
x = initial position

I've tried and tried but the answer I get is 4811.805000000001, but apparently the correct answer is -490.5m.

This is my attempt:

public class GravityCalculator2 {
    public static void main(String[] args) {
        double gravity = -9.81;
        double fallingTime = 10;
        double initialVelocity = 0.0;
        double initialPosition = 0.0;
        double x;
        x = (0.5 * ((gravity * fallingTime) * (gravity * fallingTime)) 
            + (initialVelocity * fallingTime) + (initialPosition));
        System.out.println(x);
    }
}

What did I do wrong?

Answer 1

You're code is evaluating this equation:

x(t) = 0.5 * (at)² + v(t) + x

when it should be evaluating:

x(t) = 0.5 * a(t²) + v(t) + x

The normal understanding of math notation is that exponentiation binds tighter than multiplication.