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

Question

Dr. Abubacker · Accepted Answer

(b) Let the stone remains in air for $t\ s$. From $S = ut + \frac{1}{2}gt^{2}$
Here, $u = 0,\ \ S = \frac{1}{2}gt^{2}$
Total distance travelled by the stone in last second is
$D = S_{t} - S_{t - 1} = \frac{1}{2}gt^{2} - \frac{1}{2}g(t - 1)^{2}$
Distance travelled by the stone in first three seconds is
$S_{3} = \frac{1}{2} 	imes g 	imes 3^{2} = \frac{9}{2}g$
According to given problem, $D = S_{3}$
$	herefore\frac{g}{2}(2t - 1) = \frac{9}{2}g$ or $2t - 1 = 9\  \Rightarrow t = 5s$

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

Explanation

Reviewed by Dr. Abubacker

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