Question

A stone falls freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first 3 seconds of its motion . The stones remains in the air for a. 3 sec b.5 sec c.7sec d.4 secs

Answer 1

Last sec Sn = a/2 (2n - 1) = 5(2n-1)
S in 3s= 1/2 .10.9 = 45

10n - 5 = 45
n = 5 s

Answer 2

Let us assume g to be $10\frac { m }{ { s }^{ 2 } }$.

So distance covered in first 3 secs can be calculated from this formula; $s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

Since initial velocity is 0;

$s=\frac { 1 }{ 2 } a{ t }^{ 2 }\quad \Rightarrow s=\frac { 1 }{ 2 } \times 10\times 3\times 3=45m$

Now for the last sec $s=\frac { u+a }{ 2(2n-1) }$ since this is a free fall there is no initial velocity

Hence u=0 again, that gives $s=\frac { 10 }{ 2(2n-1) }\quad \Rightarrow 45=5(2n-1)\quad \Rightarrow n=5s$