physic question

Question

"That's one small leap for man..." In a more dramatic version of his 1969 moon landing, Neil Armstrong eschews the ladder and instead leaps from the hatch of the Lunar Excursion Module to the lunar surface below. If he leaps horizontally from the hatch at 2 m/s and the hatch is 3 m above the lunar surface, with what velocity does he impact the surface? Note: gravity on the moon is 1/6 as strong as gravity on earth, causing an acceleration of only 1.6 m/s². Include a sketch that shows Neil's trajectory as a dotted line, and labeled vectors for ∆y, v_i, and v_f.

Answer 1

Given

Horizontal Initial Velocity: Vox = 2 m/s

Initial Vertical Position: Yo = 3 m

Gravity = 1/6 * g = 1/6 * 9.81 = 1.635 m/s^2

Find:

Let position at impact be labeled as point 1

Therefore, we need to find V1. Velocity at impact

The velocity at impact has 2 components V1x & V1y

Solution:

Velocity in the X direction is constant since there is no horizontal acceleration

Vox = V1x = 2 m/s

Velocity in the Y direction changes throughout the trajectory since there is gravity pushing down

2 * ay * (Y1-Yo) + Voy^2 = V1y^2

2 * (-1.635) * (0- (-3)) + 0 = V1y^2

Vy1^2 = 9.81

Vy1 = 3.132 m/s

Now,

V1 = (V1x^2+V1y^2)^1/2

V1 = (2^2 + 3.132^2)^1/2 = 3.716 m/s

Solution:

V1 = 3.716 m/s (This is equivalent to Vf)

Please let me know if you need help guiding you through creating your sketch!