Question

A particle falling under gravity travels 25 m in a particular second. Find the distance travelled by it in the next 3 seconds.

Given : Particle falls 25 m in a particular second
Sn=-25 m
u=0 m/s
To find : Distance travelled by it in next 3 seconds

Answer 1

Distance travelled by the particle in nth second is
Sn = u + `1/2` 𝑎 (2𝑛−1)
`-25 = 0- 1/2 `x 9.81 x (2n-1)
5.0968 = 2n-1
n = 3.0484
Considering n as an integer
n = 3 s
Using kinematical equation : `s= ut + 1/2at^2` ……………..(1)
S is the displacement of the particle in 3 seconds
S = 0 -` 1/2 xx9.81 xx 3^2`
S = -44.145 m
V is the displacement of particle in 6 seconds is
V = 0 -`1/2 xx 9.81 xx 6^2` (From 1)
=-176.58 m
The distance travelled by particle in `4^(th),5^(th) and 6^(th)` seconds = 176.58-44.145
=132.435 m
The distance travelled by particle in next 3 seconds is 132.435 m