Question #4fa55

1 Answer
Ujjwal
Jan 19, 2018

See below

Explanation:

Using the second equation of motion,
# s = ut + 1/2 at^2#
Put s = h
u = 0
# a = 10ms^-2#
We get,
# h = (0)t + 1/2*10*t^2#
# h = 0 + 5t^2#
# h = 5t^2# ----(1)
Thus we get a relation between distance covered by the body and time taken.
Now put # t = t/2# in the equation (1)
#h' = 5(t/2)^2#
#h' = (5t^2)/4#

#h':h = (5t^2)/4 : 5t^2#
#h':h = 1:4#

Thus, the distance covered in #1/2 t # is equal to #1/4#h.
So, the distance of the body from where it was released is #1/4#h.

Therefore,height of the body from ground is #3/4#h.

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