A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first (5 ~s ). It is in air for....... ( s ).

Step-by-step solution

Since, stone falls freely, hence its initial velocity, (u=0 ) Distance covered by the stone in first (5 ~s ) is given as ( aligned h₁ & =u t+ 1 2 g t^2=0 5+ 1 2 9.8 5^2 & =122.5 ~m aligned ) Since, distance travelled in last second ( (n )th second) is equal to distance travelled in first (5 ~s ), hence ( aligned & h₁ =u+ 1 2 g(2 n-1) & 122.5 =0+ 1 2 9.8(2 n-1) & 2 n-1 = 122.5 2 9.8 & 2 n-1 =25 & 2 n =25+1 & 2 n =26 n=13 aligned )