Question

Determine how much farther a person can jump on the moon as compared to the earth if the takeoff speed and angle are the same. The acceleration due to gravity on the moon is one sixth what it is on earth

Answer 1

Question #53790, Physics / Mechanics | Kinematics | Dynamics |

Determine how much farther a person can jump on the moon as compared to the earth if the takeoff speed and angle are the same. The acceleration due to gravity on the moon is one sixth what it is on earth

Solution:

Let's make an assumption that an object are thrown with initial velocity of $\mathbf{v}_0$ and angle of $\alpha$ above the horizontal level. Then horizontal (x) and vertical (y) velocities (x direction - parallel to ground level and y - perpendicular to ground level) are:

$$v _ {x} = v _ {x o} = v _ {0} \times \cos (\alpha)$$ $$v _ {y o} = v _ {0} \times \sin (\alpha)$$

The time, when the object riches the maximal height (y velocity component equals zero), can be found:

$$v _ {y} = v _ {y o} - a t _ {u p} = 0$$ $$t _ {u p} = v _ {y o} / a$$

Thus, total time in the air is two times longer: $t = 2t_{\mathrm{up}} = [2v_0 \times \sin(\alpha)] / a$

The maximal distance is determined due to the equation:

$$x = v _ {x} \times t = v _ {0} \times \cos (\alpha) \times [ 2 v _ {0} \times \sin (\alpha) ] / a = v _ {0} ^ {2} \times \sin (2 \alpha) / a, \text { which means x proportional to 1 / a}$$

Taking into the acceleration on the moon is of $a_{\mathrm{m}} = 1 / 6$ $a_{\mathrm{e}}$ , a ratio between maximal distances on the moon $(x_{\mathrm{m}})$ and on the earth $(x_{\mathrm{e}})$ equals:

$$x _ {m} / x _ {e} = a _ {e} / a _ {m} = 6$$

Answer: The maximal distance on the moon is 6 times longer than on the earth for the lunched object with the same take off speed and angle.