If a person can jump a maximum horizontal distance (by using a 45° projection angle) of 2.07 m on Earth, what would be his maximum range on the Moon, where the free-fall acceleration is g/6 and g = 9.80 m/s2?
The classical mechanics equation for distance traveled by a projectile, launched at speed v, angle θ, and height above the ground h, is:
d = v cos(θ) (v sin(θ)+√(v^2 sin^2(θ)+2 g h))/g
When h = 0 and θ=π/4 this reduces to simply:
d = v^2/g
Thus maximum distance is inversely proportional to gravitational acceleration. For the same initial velocity and launch angle, the distance on the moon is then: 6 * 2.07[m].
43 m/s
To find the maximum range on the Moon, we can use the range formula for projectile motion:
Range = (velocity^2 * sin(2θ)) / g
Given:
- Initial altitude (y₀) = 0 (assuming the person jumps from the ground)
- Projection angle (θ) = 45°
- Maximum horizontal distance on Earth (Rangeₑ) = 2.07 m
- Acceleration due to gravity on the Moon (gₘ) = 9.80 m/s² / 6 = 1.63 m/s²
Step 1: Determine the initial velocity (v₀) on Earth.
Using the range formula for Earth:
Rangeₑ = (v₀² * sin(2θ)) / g
Rearranging the formula, we can solve for v₀:
v₀ = sqrt((Rangeₑ * g) / sin(2θ))
Plugging in the given values:
v₀ = sqrt((2.07 * 9.80) / sin(2 * 45°))
v₀ ≈ 5.95 m/s
Step 2: Determine the new maximum range on the Moon (Rangeₘ).
Using the range formula:
Rangeₘ = (v₀² * sin(2θ)) / gₘ
Plugging in the values we calculated:
Rangeₘ = (5.95² * sin(2 * 45°)) / 1.63
Rangeₘ ≈ 8.66 m
Therefore, the person's maximum range on the Moon, where the free-fall acceleration is g/6, is approximately 8.66 meters.
To find the maximum range on the Moon, we need to consider the projectile motion equation for horizontal distance:
Range = (Initial Velocity^2 * sin(2θ)) / g
Given that the maximum horizontal distance on Earth is 2.07 m, we can find the initial velocity using the projection angle of 45°.
First, let's find the initial velocity on Earth:
Range = 2.07 m
θ = 45°
g = 9.80 m/s^2
Rearranging the equation, we get:
Initial velocity^2 = (Range * g) / sin(2θ)
Substituting the values, we have:
Initial velocity^2 = (2.07 * 9.80) / sin(90°)
= 20.266
Taking the square root of both sides, we get:
Initial velocity = sqrt(20.266)
= 4.50 m/s
Now, let's find the maximum range on the Moon.
g (gravity on the Moon) = g/6
g (gravity on the Moon) = (9.80 m/s^2) / 6
= 1.633 m/s^2
Using the same formula, but now with the gravity on the Moon, we have:
Range = (Initial Velocity^2 * sin(2θ)) / g
Range = (4.50^2 * sin(90°)) / 1.633
Since sin(90°) is equal to 1, we simplify further:
Range = (4.50^2) / 1.633
Range = 20.25 / 1.633
Range = 12.39 m
Therefore, the maximum range of the person's jump on the Moon, where the free-fall acceleration is g/6, would be approximately 12.39 meters.