To solve this problem, we will use the formula for the maximum range of a projectile, which is given by R = (v^2 * sin(2Θ)) / g. Here, v is the launch speed, Θ is the launch angle, and g is the acceleration due to gravity.
Step 1: Write down your known values: the launch speed (v) is 20.3 m/s, the launch angle (Θ) is 43.3 degrees, and g (the acceleration due to gravity) is 9.8 m/s^2.
Step 2: Before we use the formula, it's important to note that the trigonometric functions in this formula work with angles in radians, not degrees. We have to convert the angle from degrees to radians. To convert degrees to radians, we multiply by pi and divide by 180. So, our angle in radians is 43.3 * pi / 180.
Step 3: Now we can use the formula for the maximum range of a projectile: R = (v^2 * sin(2Θ)) / g. Plugging in our known values, we get R = (20.3 m/s)^2 * sin(2 * 43.3 degrees) / 9.8 m/s^2.
Step 4: Make the calculation. The solution for the maximum range of the ball is approximately 41.98 metres. Therefore, this ball you threw will travel this distance horizontally before returning to the same level as the launch height. The maximum range refers to the horizontal distance traveled by the projectile, and this is the result you were looking for.