a body falls freely from rest. it covers as much distance in the last second if its motion as covered in the first three seconds . the body has fallen for a time of?
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acceleration due to gravity=g
let the time of falling =t
velocity at t-1 sec =g×(t-1)
distance covered in first =distance covered in
3secs the last sec
0×3+1/2g(3^2)=g (t-1)×1+1/2g (1^2)
1/2g×9=g (t-1)+1/2g (diving both sides by g)
so,
1/2×9=(t-1)+1/2
1/2×8=(t-1)
4=t-1
5=t
the ball has fallen for 5 secs
let the time of falling =t
velocity at t-1 sec =g×(t-1)
distance covered in first =distance covered in
3secs the last sec
0×3+1/2g(3^2)=g (t-1)×1+1/2g (1^2)
1/2g×9=g (t-1)+1/2g (diving both sides by g)
so,
1/2×9=(t-1)+1/2
1/2×8=(t-1)
4=t-1
5=t
the ball has fallen for 5 secs
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