Question

A person can jump a horizontal distance of 1.31 m on the Earth. The acceleration of gravity is 12.3 m/s^2.

a) How far could he jump on the Moon, where the free-fall acceleration is 0.258g Answer in units of m.

b) Repeat for Mars, where the acceleration due to gravity is 0.293g. Answer in units of m.

Answer 1

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

!

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(

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^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

b) On Mars, where the acceleration due to gravity is 0.293g:

Similarly, we need to find the acceleration due to gravity on Mars:

g_Mars = 0.293g_Earth

g_Mars = 0.293(12.3 m/s^2)

g_Mars = 3.61 m/s^2

Using the same formula and rearrangement as in part a, we can find the distance he could jump on Mars:

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.61 m/s^2) / (12.3 m/s^2)

R = 0.223 m

Therefore, he could jump approximately 0.223 m on Mars.