A person can jump a maximum horizontal
distance (by using a 45� percent 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
I believe that if the acceleration decreases by a factor of 1/6, then the time will increase by a factor of 6. Therefore, the distance should be 18m.
To find the maximum range on the Moon, we need to determine the projectile range formula in terms of the given variables. The range of a projectile can be calculated using the equation:
Range = (velocity^2 * sin(2θ)) / g
Where:
- Range represents the horizontal distance traveled by the projectile
- velocity is the initial velocity of the projectile
- θ (theta) is the launch angle of the projectile
- g is the acceleration due to gravity
Let's plug in the values for Earth conditions and calculate the range on Earth:
Given:
- velocity on Earth, ve = unknown (to be determined)
- θ on Earth, θe = 45 degrees
- g on Earth, ge = 9.8 m/s^2
- Range on Earth, Re = 3 m
Using the formula, we have:
Re = (ve^2 * sin(2θe)) / ge
Simplifying the equation, we find:
3 = (ve^2 * sin(90°)) / 9.8
Since sin(90°) is equal to 1, the equation further simplifies to:
3 = ve^2 / 9.8
Rearranging the equation to solve for ve^2, we get:
ve^2 = 3 * 9.8
ve^2 = 29.4
Taking the square root of both sides to isolate ve, we find:
ve ≈ √29.4
ve ≈ 5.42 m/s
Now we have the velocity on Earth, ve.
To find the maximum range on the Moon, we need to determine the new acceleration due to gravity, gm. Given that gm is one-sixth (1/6) of ge, we have:
gm = ge / 6
gm ≈ 9.8 m/s^2 / 6
gm ≈ 1.63 m/s^2
Now, let's calculate the maximum range on the Moon, Rm, using the same formula but with the new values:
Given:
- velocity on the Moon, vm = ve (since the velocity remains the same)
- θ on the Moon, θm = 45 degrees
- g on the Moon, gm = 1.63 m/s^2
- Range on the Moon, Rm = unknown (to be determined)
Using the formula, we have:
Rm = (vm^2 * sin(2θm)) / gm
Plugging in the values, we get:
Rm = (5.42^2 * sin(90°)) / 1.63
Since sin(90°) is equal to 1, the equation simplifies to:
Rm = (5.42^2) / 1.63
Calculating this, we have:
Rm ≈ 18.02 m
Therefore, the maximum range on the Moon, when considering a 45-degree projectile angle, is approximately 18.02 meters.