Question

12. A stone is dropped on the surface of the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s . Determine the time for the stone to fall to the surface of the moon.​

Answer 1

Answer:

1.3 sec

Explanation:

kinetic energy = gravitational potential energy

½ mv² = mgh

v²= 2gh

v² = 2×1.67×1.4

v = √(2×1.67×1.4)

v= 2.16

a = v-u/t

1.67 = 2.16/t

t= 2.16/1.67

t= 1.29

t= 1.3 sec

Answer 2

Answer:

$answer$

Given, acceleration a=1.67m/s

Distance covered by feather d=1.40m

Initial velocity u=0m/s (assume heather falls from rest)

Using formula S=ut+at

2

/2,

1.40=0(t)+

21

(1.67)t 2

or t 2

=2.8/1.67=1.68

⟹ t= 1.68

=1.29s