Question

Two stones from the same height is thrown Downward at the interval of two seconds. At what time will the first Stone would be separated by other stone by 20m. [Take g :-10m/s]​

Answer 1

Answer:

ANSWER

v=u+at  or v=9.8t

t=1030=3.0s is the time at which stone will reach 30 m/s velocity 

First stone will reach s=ut+(1/2)at2 

so s1=1/2×10×9m

s1=45m 

(initial velocity = u = 0) 

second stone was dropped after 2sec, so time for itt1=3.0−2.0s=1.0s

s2=1/2×10×12=5.0m 

so distance between two stones 

D=45.0−5.0=40.400m 

Answer 2

$\mathcal{\huge{\fbox{\red{Question\:?}}}}$

✴ Two stones from the same height is thrown Downward at the interval of two seconds. At what time will the first Stone would be separated by other stone by 20m. [Take g :-10m/s]

$\mathcal{\huge{\fbox{\green{Answer:-}}}}$

➡ At distance of 2 seconds will the first Stone would be separated by other stone by 20m.

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

$\mathtt{\huge{\underline{\red{Given:-}}}}$

• Two stones from the same height is thrown Downward at the interval of two seconds.

• The first Stone would be separated by other stone by 20m.

$\mathtt{\huge{\underline{\green{To\:Find:-}}}}$

• At what time will the first Stone would be separated by other stone by 20m.

$\mathtt{\huge{\underline{\purple{Calculation:-}}}}$

• Two stones from the same height is thrown downward at the interval of two seconds.

According to the question,

Using, v = s/t

The velocity is the time rate of change of displacement. If s is the displacement of an object in some time t, then the velocity is equal to, v = S/T. The units of velocity are m/s or km/hr.

Where,

• v = 10 m/s

• s = 20 m

We know that, t = s/v

➡ t = 20/10

➡ t = 2 s

The time is 2s at which taken stone to reach 10 m/s velocity .

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