A person can jump a horizontal distance of 3.10 m on Earth, where the free-fall acceleration, g = 9.81 m/s2.

(a) How far could the person jump on the moon, where the free-fall acceleration is g/6
m
(b) How far could the person jump on Mars where the acceleration due to gravity is 0.38g?
m

To answer these questions, we can use the concept of projectile motion. The horizontal distance covered by an object in projectile motion can be calculated using the formula:

d = v * t

where d is the distance, v is the initial horizontal velocity, and t is the time of flight.

In this case, since the question gives us the distance on Earth and asks about the distance on the Moon and Mars, we need to find the time of flight and the initial horizontal velocity on both the Moon and Mars. Let's solve each part step by step:

(a) How far could the person jump on the moon?
On the moon, the free-fall acceleration is g/6. Since the horizontal motion is not affected by gravity, the initial horizontal velocity remains the same. Let's assume the initial horizontal velocity on Earth is v_E.

On the Moon:
Horizontal distance (d_Moon) = v_E * t_Moon

To find the time of flight on the Moon (t_Moon), we need to use the equation for the vertical motion:

h = (1/2) * g * t^2

The initial vertical velocity is 0 since the person is jumping horizontally. The height of the jump on the Moon is the same as on Earth, which we'll call h_E. Plugging in the values, we have:

h_E = (1/2) * g * t_E^2

And on the Moon:

h_Moon = (1/2) * (g/6) * t_Moon^2

Since the height of the jump is the same on both Earth and the Moon, we can equate the above two equations:

h_E = h_Moon

(1/2) * g * t_E^2 = (1/2) * (g/6) * t_Moon^2

Simplifying, we get:

t_Moon = √(6 * t_E)

Substituting the value of t_Moon in the horizontal distance equation, we have:

d_Moon = v_E * √(6 * t_E)

We know the distance on Earth (d_E) is 3.10 m, so we need to find the value of t_E. We can use the formula for the time of flight in vertical motion:

h_E = (1/2) * g * t_E^2

Plugging in the values, we have:

(1/2) * (9.81 m/s^2) * t_E^2 = h_E

Simplifying, we get:

t_E^2 = (2 * h_E) / g

Substituting the values, we get:

t_E^2 = (2 * 3.10 m) / 9.81 m/s^2

Solving for t_E, we have:

t_E = √(2 * 3.10 / 9.81)

Now, we can substitute the value of t_E into the equation for d_Moon to find the distance on the Moon.

(b) How far could the person jump on Mars?

On Mars, the acceleration due to gravity is 0.38g. Let's assume the initial horizontal velocity on Earth is v_E.

On Mars:
Horizontal distance (d_Mars) = v_E * t_Mars

To find the time of flight on Mars (t_Mars), we again need to use the equation for the vertical motion:

h = (1/2) * g * t^2

The height of the jump on Mars is the same as on Earth (h_E). Plugging in the values, we have:

h_E = (1/2) * g * t_E^2

And on Mars:

h_Mars = (1/2) * (0.38g) * t_Mars^2

Since the height of the jump is the same on both Earth and Mars, we can equate the above two equations:

h_E = h_Mars

(1/2) * g * t_E^2 = (1/2) * (0.38g) * t_Mars^2

Simplifying, we get:

t_Mars = √(2 * t_E / 0.38)

Substituting the value of t_Mars in the horizontal distance equation, we have:

d_Mars = v_E * √(2 * t_E / 0.38)

Again, we need to find the value of t_E. Using the same formula for the time of flight in vertical motion:

h_E = (1/2) * g * t_E^2

Plugging in the values, we have:

(1/2) * (9.81 m/s^2) * t_E^2 = h_E

Simplifying, we get:

t_E^2 = (2 * h_E) / g

Substituting the values, we get:

t_E^2 = (2 * 3.10 m) / 9.81 m/s^2

Solving for t_E, we have:

t_E = √(2 * 3.10 / 9.81)

Now, we can substitute the value of t_E into the equation for d_Mars to find the distance on Mars.

Remember to always double-check your calculations and units to ensure accuracy.