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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 .
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
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
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.
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.
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