Question

A man holds four balls 180 m above the ground
and drops them at regular intervals of time so
that when the first ball hits ground, the fourth
ball is just leaving his hand. At this time, the
second and third balls from the ground arę at
the positions-
(a) 160 m and 100 m respectively
(b) 80 m and 20 m respectively
(c) 20 m and 80 m respectively
(d) 100 m and 160 m respectively​

Answer 1

Answer:

Explanation:

Time taken by first ball h reach the ground

=2Hg−−−√=2×18010−−−−−√=6sec

time interval between two balls in 2 seconds

H1=12g(4)2=80m

H1=12g(2)2=20m

Height from ground are 180−80=100m

180−20=160m

Hence d is the correct answer.

Answer 2

Answer:

First ball from ground = 100m

Second ball from ground = 160m

Given:

A man holds four balls 180 m above the ground and drops them at regular intervals of time so that when the first ball hits ground.

To find:

Find the position of second and fourth ball from the ground , if at that position  the first ball hits ground, the fourth ball is just leaving his hand. At this time, the

Explanation:

As all the balls leave the hand at some interval of time , so we have to find the time of interval.

time taken by first ball to reach tha ground.

S = ut + 1/2gt²

S = 1/gt² ( u = 0)

(180 x2)/10 = t²

t = √36

t=6 sec

so time taken to reach the ground in 6 seconds

therefore all  the three balls will be allowed to fall after every 6/3 seconds = 2 seconds

When the first ball reaches the ground.

1. second ball would have covered 4 seconds, so we will find the distance in 4 seconds,

S = 1/2gt²

  =1/2x10x4²

  =80 m

so the distance of second ball  from the ground will be 180- 80 = 100 m

2.Third ball second ball would have covered 2 seconds, so we will find the distance in 2 seconds,

S= 1/2x10x2²

  =20

so the distance of third ball  from the ground will be 180- 20 = 160 m