Question

the feather is dropped on the Moon from the height of 1.40 meters. the acceleration of gravity on the moon is 1.67 M per second square.
determine the time for the feather to fall to the surface of the Moon

Answer 1

u=0
a=1.67
s=1.4

$v {}^{2} + = {u }^{2} + 2as$
hence v2 is 4.676
v is 68.2
time taken is
$s = ut + 1 \div 2at {}^{2}$
1.4=1/2 1.67 t2
2.8/1.67=1.6766
t= root of 1.6766
t=1.29 sec