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# A ball is dropped from a height .If it takes0.200 second to cross the last 6.00 m before hitting the ground ,find the height  from which it was dropped .

lokesh soni
37 Points
8 years ago

6=g*t2/2-g(t-0.2)2/2

6=10/2(t2-t2+0.4t-0.04)

6/5+0.04=0.4t

t=3.1s

s=gt2/2

s=10/2*(3.1)2

s=48.05m

Prajwal kr
49 Points
8 years ago

We need to find H

Using second eqn of motion:

6=0.2u-5(0.04).............................u is velocity at h=6m

u=31m/s

Conserving energy:

mgh=0.5m(31)2.......................................Datum line for P.E is at 6 m from ground. h= H-6

=>h=48.05m

=>H=54.05m

naninagaraju sivacharan lakshmi narsing
34 Points
8 years ago

we consider a ball it is dipped from a height equal to square of the intiial distance by 2 multiplide with g(9.8),

i.e ans is 1.83 metres.

Akash Kumar Dutta
98 Points
8 years ago

total height h=1/2.g.t^2

for (t - .2) secs it travels (h - 6) m

so h-6 = 1/2.g.(t - .2)^2

putting h=1/2.g.t^2 and solving we get

t=3.09 secs.

so h=1/2.g.(3.09)^2

h=47.75 m  (ANS)

hope i helped you.