Q.2 An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height of 100 m. What is the difference in their heights after 2 s if both the objects drop with the same accelerations? How does the difference in heights vary with time?
Dear Student,
Let
h1 = 150 m and h2 = 100 m
and we know
s = ut + (1/2)at2
or as in this case, u =0 and a = -g
so
s = -(1/2)gt2
or
s1 = s2 = -(1/2)gt2 = -(1/2)x10x(2)2 = 20 m
which is the distance travelled by the two objects.
now the height of the object 1 after 2 seconds will be
h'1 = 150 - 20 m = 130 m
and similarly for object 2
h'2 = 100 - 20 = 80 m
so, the difference in heights will still be = 130 - 80 m =50m
and their difference in height would stay constant until object 2 touches the ground.
Regards,