Question

Two particles are released from the same height at an interval of 1 second how long after the first particle begins to fall will the two partocles be 10 m apart

Answer 1

Notice that each particle moves zero initial velocity and constant acceleration
g = 10 m/s².
Hence the distance passed by the first particle at time t is given by the formula:
h1(t) = gt²/2

The second particle is released at t = 1 s, so the distance passed by the second particle at time
t ≥ 1 s is given by the formula:
h2(t) = g(t − 1)²/2


We should find t such that
h1(t) − h2(t) = 10 m.
Thus
gt²/2 − g(t − 1)²/2= 10
g/2(t² − (t − 1)²) = 10
t² − t² + 2t − 1 = 10 *2/g
2t =( 10 *2/g)+ 1
t =10/g+1/2
Substituting g = 10 m/s²
we obtain that
t =
10/10+1/
2
= 1 + 0.5 = 1.5 s.