When the Apollo 11 lunar module Eagle lands on the moon it comes to a stop 10m above the surface of the moon. The last 10m it freely falls to the surface of the Moon. i) How long does it take for the Eagle to touch down?
ii) What is the velocity of the lunar module when it hits the surface of the moon?

i) 3.514 s, ii) 5.692 m/s

Explanation:

i) We can use Newton's second law of motion to find out how long does it take for the Eagle to touch down.

as the equation says for free-falling

h = ut +0.5gt^2

Here, h = 10 m, g = acceleration due to gravity = 1.62 m/s^2( on moon surface)

initial velocity u = 0

10 = 0.5×1.62t^2

t = 3.514 seconds

Therefore, it takes t = 3.514 seconds for the Eagle to touch down.

ii) use Newton's 1st equation of motion to calculate the velocity of the lunar module when it hits the surface of the moon

v = u + gt

v = 0+ 1.62×3.514

v= 5.692 m/s

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

less than one you can't create energy