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Displacement – A Numerical Problem

A fly is sitting in the middle of a wall of a cubic room of side ’a’.If it flies and sit to one of opposite corner of the wall. Find its displacement?

(Harmeet Singh asked)
Answer:
If the fly sit on the same wall on the horizontal, (from A to B in the diagram) the displacement is “a
If the fly sits on the diagonally opposite corner on the same wall, (from A to C in the diagram) the displacement is “√2 a
If the fly sits on the corner on the longest diagonal of the room,  (from A to H in the diagram) the displacement is “√3 a

Problem from Kinematics

Devika posted:

There are two particles A and B moving with same velocity u such that initially there velocity are perpendicular to each other and distance between them is L.the velocity of A is always directed towards B.find the distance between them after a long time.

Ans will be posted soon

Ball falls to earth, and earth moves to ball?!!

Sandeepan Dhar Choudhary asked:

“When the ball falls on the earth, the earth also moves up to meet it. But the motion of the earth is not noticeable. Explain why?”

Answer:

The force exerted by the ball on earth is same as the force exerted by the earth on the ball. The acceleration of the ball towards earth is approximately 10 m/s2.

The acceleration of earth towards the ball = force/mass of earth ≈ 10-23 m/s2

When the earth moves towards the ball with this acceleration, the time required to travel the distance will be very large. If we assume a distance of 1m, and from the equation

$S=ut+\frac{1}{2}at^{2}$

we get t ≈ 1012 seconds whereas the ball reaches the earth in 0.4 seconds

[warning]In a nut shell, we can say that we cannot notice the motion of earth towards the ball due to the extremely low value of acceleration of earth towards the ball.[/warning]

Another problem from kinematics

Zeen asks:

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.800 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

Answer :

let the total height be h

So, for first case

S = h/2

a = g=10 m/s^2

u=0 m/s

t = 0.8 sec

using the relation $S=ut + \frac{1}{2}at^{2}$

h/2 = 0.5 x 10×0.8 ^2

h=6.4 m

In second case (considering the full motion)

S=h=6.4m

t=?

a=g=10m/s^2

u=o m/s

using the relation $S=ut + \frac{1}{2}at^{2}$
6.4 = 0.5 x 10 x t^2

or

t = 2 x 6.4/10 = 1.28 s

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