Question

A body falls from a certain height.Two seconds later another body falls from the same height.How long after the beginning of motion of the first body is the distance between the bodies twice the distance at the moment the second body starts to fall? Cmmon Physic masters!!!

Answer 1

initial separatn was 20m. After 3 sec it is 40m

Answer 2

The correct answer is 3 seconds.

GIVEN

A body falls from a certain height.

Another body falls after 2 seconds.

TO FIND

Time after which the distance between the two objects is twice the distance at the moment the second body falls.

SOLUTION

We can simply solve the above problem as follows.

Let,

The initial velocities of the First and second bodies are, u1 and u2 respectively.

And,

u1 = u2 = 0 (free fall)

Let the height covered by the First body in two seconds be, h.

Applying the second equation of motion,

$h = ut + \frac{1}{2} g \times 2 \times 2$

h = 2g. (equation I)

Let the time at which the distance between the two bodies is twice the distance before the second body falls be, T.

Let the first body cover a distance of h1 At time T seconds

And, let the second body covers a distance of h2 at time (T-2) seconds

ATQ,

$h1 - h2 = 2h$

h1 - h2 = 4g. (Equation 2)

Applying the second equation of motion for the first body to cover h1 height in T seconds -

$h1 = 0 \times T + \frac{1}{2} \times g \times {T}^{2}$

$h1 = \frac{1}{2}g {T}^{2}$

Similarly, Applying the second equation of motion for the second body for covering a distance of h2 in T-2 seconds;

$h2 = 0 \times (T - 2) + \frac{1}{2} \times g \times (T - 2)^{2}$

$= \frac{1}{2} g (T^{2} + 4T + 4 )$

$= \frac{1}{2} gT^{2} + 2g - 2gT$

Putting the values of h1 and h2 in equation 2;

$\frac{1}{2} gT^{2} -( \frac{1}{2}gT ^{2} + 2g - 2gT )$

$= - 2g + 2gT$

-2g + 2gT = 4g

T = 3 seconds.

Hence, The correct answer is 3 seconds.

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