Question

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!!!!!!!!!!!!!!

Answer 1

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