To find the free-fall acceleration on the planet, we can use the following equation for projectile motion:
d = (v^2 * sin(2θ)) / g
Where:
- d is the horizontal distance traveled (15.0 m in this case).
- v is the initial speed (3.00 m/s in this case).
- θ is the launch angle.
- g is the acceleration due to gravity.
In this problem, we are given the maximum horizontal distance and the initial speed. We need to find the value of the acceleration due to gravity (g). To do this, we can rearrange the equation as follows:
g = (v^2 * sin(2θ)) / d
The maximum horizontal distance occurs when the launch angle is 45 degrees (sin(90 degrees) = 1). Plugging in the given values:
g = (3.00^2 * sin(2 * 45°)) / 15.0
Simplifying the equation:
g = (9.00 * 1) / 15.0 g = 0.60 m/s^2
Therefore, the free-fall acceleration on the planet is 0.60 m/s^2.