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Dear Aditya
take downward direction as positive
Y=uyt + at2/2
put Y=100m, a=g=10 m/s2, uy=0
100=5t2
t=2√5 sec
now t/2=√5 sec
distance covered after t/2 sec
again apply the formula
and put t=√5 sec
Y=10*√5*√5/2
Y=25m
height from ground= 100-25=75m
All the best.
AKASH GOYAL
AskiitiansExpert-IITD
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body takes t sec to reach the groung..
initial velocity is zero..
s=ut +at2 /2
-100=-gt2 /2
t=2sqrt5 sec
now displacement of particle=s=gT2 /2
for T=t/2=sqrt5sec
S=25m
so the particle is at (100-25)=75m above the ground at T=t/2...
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