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aditya kulkarni Grade: 10
        if a body is droped from 100m it takes        t seconds to reach the ground.then find its height at time=t\2 seconds......
7 years ago

Answers : (3)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

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


Y=uyt + at2/2


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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7 years ago
Amit Tapas
18 Points
										See,

by 2 kinematical equation
s=ut+1/2 at(squared)
given
s=100
t=t
u=0
therefore
100=1/2*a*t(squared)


so

a=200/t(squared)

to find height
when t=t/2

s=ut+1/2at(squared)

s=1/2*200/t(squared)*t(squared)/4


s=25m

the body reaches the height of 25m at t/2 seconds

7 years ago
vikas askiitian expert
510 Points
										

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...

7 years ago
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