Question

A particle is thrown vertically upwards with velocity 30m/s .The distance covered by the particle in 9th second(g=10m/s2)

Answer 1

a particle is thrown vertically upwards with velocity 30 m/s .
so, time taken to reach maximum height ;
v =u + at
0 = 30 - 10t
t = 3sec
here in question, didn't mention about collision is elastic or inelastic . so, we have to assume body just make round trip then rest.

hence, body moves only 6 sec { 3 sec for upward direction and 3 sec for downward direction }
after that body becomes rest.
so, distance covered by particle in 9th sec = 0.

Answer 2

Given:

Velocity = 30 m/s

g = 10 m/s^2

To find:

The distance covered by the particle in 9th second

Solution:

According to the laws of motion,

v =u + at

where,

v - Final velocity

u - Initial velocity

a - acceleration due to gravity(g)

Hence,

0 = 30 - 10t

t = 3 s

The particle goes up and comes down to rest.

The time taken to go up = 3 s

The time taken to come down = 3 s

The is at rest after 6 s.

Hence, it does not move anymore.

The distance covered by the particle in 9th sec = 0.

It is at rest.