If a person can jump a maximum horizontal distance (by using a 45° projection angle) of 2.35 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?
b) (b) Repeat for Mars, where the acceleration due to gravity is 0.38g.
To find the maximum range on the Moon, we can use the projectile motion equations. The range, R, of a projectile can be calculated using the formula:
R = (v^2 * sin(2θ)) / g
where v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
Since we are given the maximum horizontal distance on Earth, we know that the range on Earth is 2.35 m. We can use this information to find the initial velocity.
First, let's find the initial velocity on Earth by rearranging the equation for range:
v^2 = (R * g) / sin(2θ)
v^2 = (2.35 * 9.80) / sin(90)
v^2 = 22.963
v ≈ 4.792 m/s (approximated to three decimal places)
Now that we have the initial velocity on Earth, we can find the range on the Moon by using the free-fall acceleration of g/6.
Substituting the values into the range formula:
R_moon = (v^2 * sin(2θ)) / (g / 6)
R_moon = (4.792^2 * sin(2 * 45°)) / (9.80 / 6)
R_moon ≈ 7.882 m (approximated to three decimal places)
Therefore, the maximum range on the Moon would be approximately 7.882 meters.
To calculate the maximum range on Mars, where the acceleration due to gravity is 0.38g, we can use the same approach as above.
R_mars = (v^2 * sin(2θ)) / (0.38g)
R_mars = (4.792^2 * sin(2 * 45°)) / (0.38 * 9.80)
R_mars ≈ 64.734 m (approximated to three decimal places)
Therefore, the maximum range on Mars would be approximately 64.734 meters.