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
d = 6*3m = 18 m.
To find the maximum range on the Moon, we need to use the principles of projectile motion. The maximum range occurs at a launch angle of 45 degrees. The formula for the range of a projectile is given by:
Range = (velocity^2 * sin(2θ)) / g
Where:
- Range is the horizontal distance traveled by the projectile
- Velocity is the initial velocity of the projectile
- θ is the launch angle
- g is the acceleration due to gravity
On Earth, the acceleration due to gravity is 9.8 m/s^2. Given that the person can jump a maximum horizontal distance of 3 m on Earth, we can use this information to find the initial velocity.
Rearranging the formula for range, we can solve for velocity:
Velocity = √((Range * g) / sin(2θ))
Substituting the values:
Range = 3 m
g = 9.8 m/s^2
θ = 45 degrees = π/4 radians
Velocity = √((3 * 9.8) / sin(2 * π/4))
Velocity = √(29.4 / sin(π/2))
Velocity = √29.4
Now, to find the range on the Moon, we need to consider the different acceleration due to gravity. On the Moon, the free-fall acceleration is g/6. We can use this value to calculate the new range using the same formula.
Range on the Moon = (Velocity^2 * sin(2θ)) / (g/6)
Substituting the known values:
Velocity = √29.4
θ = 45 degrees = π/4 radians
g (on the Moon) = (9.8 m/s^2) / 6 = 1.633 m/s^2
Range on the Moon = (29.4 * sin(2 * π/4)) / 1.633
Range on the Moon = 29.4 / 1.633
Range on the Moon ≈ 18.02 m
Therefore, the maximum range the person can jump on the Moon, with a 45-degree launch angle, is approximately 18.02 meters.