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`        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

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

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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
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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