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  4. 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 three seconds of its motion. The stone remains in the air for

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Q. 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 three seconds of its motion. The stone remains in the air for

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Solution:

Given that the stone falls from rest, hence $u=0$
Now, distance covered in the last second is equal to distance covered in the first three seconds of the motion.
Hence, $S_{3}=S_{t}$
$\Rightarrow S_{3}=0+\frac{1}{2} g t^{2}=\frac{1}{2} \times 10 \times 9=45$
and so, $S_{t}=45=u+(2 t-1) 10 / 2$
$\Rightarrow 0+5(2 t-1)=45$
$\Rightarrow t=5\, s$