Question

a stone dropped from a height covers 24.5 m in last second of its fall The time of fall is ​

Answer 1

Answer:

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Answer 2

Answer:

The time of fall is 3 sec

Explanation:

As the stone was dropped its initial velocity was 0, i.e u = 0

Suppose the gravitational acceleration be 9.8 Newton/Kg = g

Let the total time taken to fall n sec.

Therefore in n sec it falls a height as below

$h=un+\frac{1}{2}gn^2 (Where h be the height that the stone has fallen)$

In (n-1) second it has fallen a height as below

$h_1=u(n-1)+\frac{1}{2}g(n-1)^2 (Where h_1 be the height that the stone has fallen)$

In last second it has fallen = $h-h_1$

$\implies un+\frac{1}{2}gn^2 - [u(n-1)+\frac{1}{2}g(n-1)^2]\\\\\implies un+\frac{1}{2}gn^2 - un+u-\frac{1}{2}g(n^2-2n+1)\\\\\implies un+\frac{1}{2}gn^2 - un+u-\frac{1}{2}gn^2+gn-\frac{g}{2}\\\\\implies u+gn-\frac{g}{2} = gn-\frac{g}{2} (as the Initial velocity = u = 0) \\\\ According to the question we get \\gn-\frac{g}{2}=24.5\\\\9.8n-\frac{9.8}{2}=24.5\\\\9.8n=24.5+\frac{9.8}{2}=24.5+4.9=29.4\\\\n=\frac{29.4}{9.8}=3 sec$