Question

TWO bodies begin to fall freely from the same height The second one begins to fall t sec after the first The time after which the first body begins to fall the distance between the bodies equal to l is​

Answer 1

Answer:

TWO bodies begin to fall freely from the same height The second one begins to fall t sec.

Explanation:

S = 1/2 gT^2 and s = 1/2 g(T - t)^2;

We have to find  T when S - s = L = 1/2 g (T^2 - (T^2 - 2Tt + t^2))  

2L/g = 2Tt - t^2 = (2T - t)t  

2L/gt + t = 2T; T = L/gt + t/2

Answer 2

Answer:

The required time is

$T = \frac{L}{gt} + \frac{t}{2}$

Explanation:

Given that,

Two bodies begin to fall freely from the same height and the 2nd body begins to fall t sec after 1st body.

Let the time taken by first body be T.Therefore,

$H_{1}= \frac{1}{2} gT^{2}$

Therefore,the distance covered by second body is

$H_{2}= \frac{1}{2} g(t - T)^{2}$

According to given question,the distance between two bodies is L.

$H_{1}-H_{2}=L$

Substituting the distances in the above relation we get

$\frac{1}{2} gT^{2} - \frac{1}{2} g(t - T)^{2} = L \\ = > \frac{g}{2} (2tT - {t}^{2} ) = L \\ = > 2tT = \frac{2L}{g} + {t}^{2} \\ = > T = \frac{L}{gt} + \frac{t}{2}$

Therefore,the required time is

$T = \frac{L}{gt} + \frac{t}{2}$

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