(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?
m
(b) Repeat for Mars, where the acceleration due to gravity is 0.38g.
a. dmax = 6 * 3.43m = 20.58 m.
b. dmax = 3.43m/0.38 = 9.03 m.
(a) Well, if we're talking about jumping on the Moon, I guess you can say his range would be "out of this world!" With the lower free-fall acceleration on the Moon, his maximum range would actually increase. To calculate it, we can use the range formula: R = (v^2 * sin(2θ))/g. Since the projection angle is 45°, we can plug in the values and solve! So, R = (v^2 * sin(90))/g, which simplifies to R = (v^2 * 1)/g. On the Moon, g is g/6, so his maximum range would be 6 times larger compared to Earth. Therefore, the maximum range on the Moon would be 3.43 m * 6 = 20.58 m.
(b) Now, onto Mars! If this person can make jumps on Mars too, then they're definitely "jumping star!" The acceleration due to gravity on Mars is 0.38g. So, we can use the range formula again: R = (v^2 * sin(2θ))/g. Simplifying it with the values we have, R = (v^2 * 1)/0.38g. This means the maximum range on Mars would be 3.43 m * (1/0.38) = 9.02 m. So, on Mars, the maximum range would be 9.02 meters, making it an out-of-this-world jumping experience!
To find the maximum range on the Moon and Mars, we need to use the range formula for projectile motion:
R = (v^2 * sin(2θ)) / g
where:
R is the maximum range
v is the initial velocity of the projectile
θ is the projection angle
g is the acceleration due to gravity
We'll start by calculating the initial velocity (v) on Earth using the given information.
(a) On Earth:
Given:
Maximum horizontal distance (range) on Earth, R = 3.43 m
Projection angle, θ = 45°
Acceleration due to gravity on Earth, g = 9.80 m/s^2
To find the initial velocity, we rearrange the range formula:
v = sqrt((R * g) / sin(2θ))
Substituting the given values:
v = sqrt((3.43 * 9.80) / sin(2 * 45°))
Using trigonometric identities:
v = sqrt(33.6226 / sin(90°))
v = sqrt(33.6226)
Therefore, the initial velocity on Earth is approximately 5.8 m/s.
Now we can calculate the maximum range on the Moon and Mars.
(b) On the Moon:
Acceleration due to gravity on the Moon, g_moon = g / 6
Using the range formula:
R_moon = (v^2 * sin(2θ)) / g_moon
R_moon = (5.8^2 * sin(2 * 45°)) / (9.80 / 6)
R_moon = (33.64 * sin(90°)) / 1.63
Since sin(90°) equals 1:
R_moon = 33.64 / 1.63
Therefore, the maximum range on the Moon is approximately 20.65 m.
(c) On Mars:
Acceleration due to gravity on Mars, g_mars = 0.38 * g
Using the range formula:
R_mars = (v^2 * sin(2θ)) / g_mars
R_mars = (5.8^2 * sin(2 * 45°)) / (0.38 * 9.80)
R_mars = (33.64 * sin(90°)) / 3.724
Since sin(90°) equals 1:
R_mars = 33.64 / 3.724
Therefore, the maximum range on Mars is approximately 9.04 m.