Question

plz solve q no 40

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Answer 1

Answer:-

Option → C

Given :-

A body is dropped from a height under acceleration due to gravity.

If t1 and t2 is the time for first half distance and second half distance.

To find :-

The relation between $t_1 \: and \: t_2$

Solution:-

Let the time taken to reach the body in ground be T.

Then,

$T = t_1 + t_2$

Now,

u = 0

a = +g

From equation of motion under gravity :-

$\huge \boxed {h = uT + \dfrac{1}{2}gT^2 }$

→$h = 0 \times T + \dfrac{1}{2}gT^2$

→$h = \dfrac{1}{2}\times g (t_1 + t_2 )^2$

As $T = t_1 + t_2$

→$\dfrac{2h}{g} = (t_1 + t_2 )^2 -----1$

For covering half distance in time t1.

$\dfrac{h}{2} = ut_1 +\dfrac{1}{2}g(t_1)^2$

→$\dfrac{h}{2}= \dfrac{1}{2}g(t_1)^2$

→$h = g(t_1)^2------2$

Now, from equation 1 and 2,

→$\dfrac{2 \times g(t_1)^2}{g}= (t_1 + t_2)^2$

$2(t_1)^2 = (t_1 + t_2)^2$

→$(t_1)^2 = \dfrac{(t_1+t_2)^2}{2}$

→$t_1 = \sqrt{\dfrac{(t_1+t_2)^2}{2}}$

→$t_1 = (t_1+t_2) \times 1.414$

→$1.414t_1 = t_1 + t_2$

→$1.414(t_1 - t_1) = t_2$

→$t_1 ( 1.414 - 1) = t_2$

→$t_1 0.414 = t_2$

→$t_1 =t_2 \dfrac{1}{0.414}$

→$t_1 = 2.41 t_2$

hence, the relationship between t_1 and t_2 is $t_1 = 2.41 t_2$

Answer 2

Answer—

$\bf\huge\boxed{t1=2.414t2}$

SOLUTION —

SEE ATTACHMENT FOR DETAILED EXPLANATION ✌

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