(a) If a person can jump a maximum horizontal distance (by using a 45° projection angle) of 3.43 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 distance he can cover in a long jump is (Vo^2/g). If he could jump at the same velocity Vo, he would travel six times farther, because of the lower value of g.
It isn't that simple however. Space suits would have to be worn. You can run at a higher speed in a track suit on Earth than a space suit of the moon.
To find the maximum range on the Moon, we can use the range equation for projectile motion, which is given by:
range = (2 * velocity^2 * sin(angle) * cos(angle)) / acceleration
First, let's find the velocity on the Moon. Since the maximum horizontal distance is given on Earth, we need to find the corresponding initial velocity on the Moon. The vertical component of the velocity at a 45° projection angle is the same as the horizontal component.
So, the initial velocity on Earth would be:
velocity on Earth = 3.43 m
Since the free-fall acceleration on the Moon is g/6, the acceleration on the Moon would be:
acceleration on Moon = (9.80 m/s^2) / 6
Now, let's calculate the maximum range on the Moon using the range equation:
range = (2 * velocity^2 * sin(angle) * cos(angle)) / acceleration
Substituting the values we have:
range = (2 * (3.43 m)^2 * sin(45°) * cos(45°)) / ((9.80 m/s^2) / 6)
To calculate the range, we need to convert the angle from degrees to radians:
sin(45°) = sin(45 * π/180)
cos(45°) = cos(45 * π/180)
Substituting these values:
range = (2 * (3.43 m)^2 * sin(45 * π/180) * cos(45 * π/180)) / ((9.80 m/s^2) / 6)
Now, we can solve this equation to find the maximum range on the Moon.