Question

when an object is dropped from the height of 100m from the surfaceoftheearth.after3secwhatwillbethe distancetravelledby the object

Answer 1

Apply S=ut + 1/2 at^2
since.. initial velocity is 0 ie; It is at rest when dropped...
& also... it is under free fall.. so

distance travelled in 3 second =
S=0×3+1/2×9.83×3^2
=0+1/2×9.83×9
=44.273

Now subtract the distance from 100 m
=100-44.273= 55.727 m

**Note** I 've taken value of "g"=9.83 m/s^2

Answer 2

u = 0m/s

t = 3sec

g = 9.8m/s^2

according to Newton's second equation of motion

h = ut+1/2gt^2

h = 0+1/2×9.8×9

h = 4.5×9

h = 40.5metres

Therefore the distance travelled by the object after 3 sec will be 40.5 metres.