To calculate the distance a projectile travels when launched at an angle with a given velocity, we can break down the motion into horizontal and vertical components and use kinematic equations. However, to solve this problem, we can apply the formula for the range of a projectile launched from and landing at the same height, which is given by:
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
where:
- [tex]\( R \)[/tex] is the range of the projectile or how far it goes,
- [tex]\( v \)[/tex] is the speed of the projectile at launch (43.3 m/s in this case),
- [tex]\( \theta \)[/tex] is the angle of launch with respect to the horizontal (12 degrees in this case), and
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately 9.81 m/s² on Earth).
The formula incorporates the factor of [tex]\(\sin(2\theta)\)[/tex] because the range is maximized at 45 degrees; since the sine function has a period of 180 degrees, this creates symmetry in projectile range for angles complementary to 90 degrees.
Let's proceed with the calculations:
1. Convert the launch angle to radians, as standard trigonometric functions in calculations use radians:
[tex]\[ \theta = 12\; \text{degrees} \times \frac{\pi \; \text{radians}}{180\; \text{degrees}} \][/tex]
[tex]\[ \theta \approx 0.2094 \; \text{radians} \][/tex]
2. Calculate the sine of double the angle:
[tex]\[ \sin(2\theta) = \sin(2 \times 0.2094) \][/tex]
[tex]\[ \sin(2\theta) \approx \sin(0.4188) \][/tex]
[tex]\[ \sin(2\theta) \approx 0.4077 \][/tex]
3. Insert the values into the range formula and solve for [tex]\( R \)[/tex]:
[tex]\[ R = \frac{43.3^2 \times 0.4077}{9.81} \][/tex]
[tex]\[ R = \frac{1874.49 \times 0.4077}{9.81} \][/tex]
[tex]\[ R = \frac{764.41}{9.81} \][/tex]
[tex]\[ R \approx 77.92 \][/tex]
Therefore, the golf ball would travel approximately 77.92 meters, assuming no air resistance and that it lands at the same height from which it was struck. Remember to round to two decimal places as requested:
[tex]\[ R \approx 77.92 \; \text{m} \][/tex]
