A person can jump a maximum horizontal
distance (by using a 45� degree projectile angle) of 3 m on Earth. The acceleration of gravity is 9.8 m/s2. What would be his maximum range on the Moon, where the free-fall acceleration is g/6 ?
Answer in units of m
To find the maximum range on the Moon, we can use the range formula for projectile motion:
Range = (initial velocity^2 * sin(2θ)) / gravity acceleration
Given that the person can jump a maximum horizontal distance of 3 m on Earth with a 45° projectile angle, we can determine the initial velocity on Earth.
At an angle of 45°, the vertical and horizontal components of the initial velocity are equal. Therefore, the initial velocity (V₀) can be found using the formula:
V₀ = 3 m / (sin(45°))
Using the given acceleration due to gravity on Earth (9.8 m/s²), we can calculate the initial velocity:
V₀ = 3 m / sin(45°) = 3 m / √(2)/2 = 3√2 m/s
Now, let's determine the maximum range on the Moon. The free-fall acceleration on the Moon is g/6, where g is the acceleration due to gravity on Earth (9.8 m/s²).
Using g/6 as the gravity acceleration and the calculated initial velocity (V₀), we can substitute these values into the range formula:
Range = (V₀^2 * sin(2θ)) / gravity acceleration
= ( (3√2 m/s)^2 * sin(2 * 45°)) / (9.8 m/s² / 6)
= ( 18 * sin(90°)) / (9.8 m/s² / 6)
= ( 18 * 1) / (9.8 m/s² / 6)
= 18 * 6 / 9.8 m
Now, let's calculate the maximum range:
Range = 18 * 6 / 9.8 m
= 108 / 9.8 m
= 11.02 m
Therefore, the person would have a maximum range of approximately 11.02 meters on the Moon.