# A stone is dropped from a cliff at 2:30:30 p.m. Another stone is dropped from the same point at2:30:31 p.m. Find the seperation between the stones at- a.​2:30:31 p.m., b.​2:30:35 p.m. plase reply fast as i have my test tomorrow!!!!!!!!!!!!!!

1
by TustiMagar

2015-09-10T23:00:06+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Let the first stone be dropped at  time t = 0.
then the second stone is dropped at time t = 1 sec.

initial velocity for both stones = 0 m/s
acceleration = g.

The distance travelled by the first stone in time t sec =
s1 = u t + 1/2 a t²
s1 = g t² /2
distance travelled by the second stone in time duration (t -1) sec.
s2 = g (t -1)² /2

the separation between them :  s1 - s2 = g/2 * [t² - (t-1)²]
= g (t - 1/2)

This can also be found using the formula for distance travelled by a particle with uniform acceleration a during the  nth second:   S_n = (n - 1/2) a

calculations:
distance of separation = 10 * (1 - 1/2) = 5 meters at 2:30:31 pm
distance of separation = 10 * (5 - 1/2) = 45 meters at 2:30:35 pm

click on thanks button above please