Question

a ball frelly from rest and tatal distance covered by it in the last second od its motion is equal to the distance coveresd by its first five secounds of its motion the total time of the stone when it is motion would be​

Answer 1

The distance travelled by the stone in the nth second is given by,

Sn=u+a/2(2n-1)

here u=o so Sn=a/2(2n-1)

if n is last second then,Sn is equals to first five seconds

S=ut+1/2a(t)2

  =1/2a25=25a/2

1/2a(2n-1)=25a/2

=by solving this we get n=13 seconds

13th second is the last second. So, the stone will remain in air for 13 seconds.

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