Question

A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the

moon is 1.67 m/s2 . Determine the time for the feather to fall to the surface of the moon.

Answer 1

v^2=u^2+2.a.s   (where v=final velocity , u=initial velocity , s=distance covered , a=acceleration)
so here u=0 ; s=1.4 ; a=1.67
=> v^2 = 0+2*1.67*1.4
            =4.676
  => v = 2.16 m/sec
now,
v=u+at
=>  t = (v-u)/a
        =(2.16-0)/1.67
        =1.294 sec