Question

Physics Lovers!!!
Genius question xD

For a body that is dropped from a height, find the ratio of the velocities acquired at the end of 1 sec, 2 sec and 3 seconds, respectively.

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Answer 1

Heya!!

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Since body is dropped from a height, it means u= 0.

Velocity acquired by body at the end of 1 sec = v1 = u + g(1) = g
=) v1 = g .. Eq1.

Velocity acquired by body at the end of 2 sec = v2 = u + g(2) = 2g
=) v2 = 2g .. Eq2.

Velocity acquired by body at the end of 3 sec = v3 = u + g(3) = 3g
=) v3 = 3g .. Eq3.

By eq 1,2 and 3 :

=) v1:v2:v3 = g: 2g : 3g

= 1:2:3

Hope it helps uh!!

Answer 2

$Hey \:  !!$

Here is your answer..

Given that, the body is dropped from a height which means the initial velocity of body is zero i.e., u = 0.

Now, the velocity that'll be acquired by body at 1 sec,

$v1 = u + gt \\ v1 = 0 + g(1) \\ v1 = g \:  -  -  -  - (i)$

Similarly, the velocity that'll be acquired by body at 2 sec,

$v2 = u + gt \\ v2 = 0 + g(2) \\ v2 = 2g \:  -  -  -  - (ii)$

Also, the velocity that'll be acquired by body at 3 sec,

$v3 = u + gt \\ v3 = 0 + g(3) \\ v 3 = 3g \:  -  -  -  -  (iii)$

So the ratio will be :

$v1 \:  : \:  v2 \:  :  \: v3  \: =  \: g \:  \:  :  \: 2g  \: :  \: 3g$

Hope it helps...