Question

Two particles A and B released from a large height at tanda 1 second respectively. Distance between two particles att 3 seconds
NO
25 m
50 m
45 m
20 m

Answer 1

Given:

Two particles A and B released from a large height at an interval of 1 second respectively.

To find:

Distance between them after 3 seconds.

Calculation:

The particle released first will actually travel for a time of 3 seconds.

$d1 = ut + \frac{1}{2} g {t}^{2}$

$= > d1 = 0+ \dfrac{1}{2} g {(3)}^{2}$

$= > d1 = 0+ \dfrac{1}{2} \times 10 \times 9$

$= > d1 = \dfrac{1}{2} \times 10 \times 9$

$= > d1 = 45 \: m$

The second particle travels for 3-1 = 2 seconds;

$d2= ut + \frac{1}{2} g {t}^{2}$

$= > d2= 0+ \frac{1}{2} g {(2)}^{2}$

$= > d2= 0+ \dfrac{1}{2} \times 10 \times {(2)}^{2}$

$= > d2= 0+ \dfrac{1}{2} \times 10 \times 4$

$= > d2= 0+ 20$

$= > d2=20 \: m$

So , distance between them

$= d1 - d2$

$= 45 - 20$

$= 25 \: m$

So, final answer is:

Distance between them is 25 m.