if a body is droped from 100m it takes t seconds to reach the ground.then find its height at time=t\2 seconds......

Dear Aditya

take downward direction as positive

Y=u_yt + at²/2

put Y=100m, a=g=10 m/s², u_y=0

100=5t²

t=2√5 sec

now t/2=√5 sec

distance covered after t/2 sec

again apply the formula

Y=u_yt + 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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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

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...