Question

A small ball is released from rest and it
accelerates due to gravity, after 1 sec suddenly
gravity disappears and again appears after
5 sec. Ball takes total 8 sec to reach the
ground. The height above ground from where
ball was released, is :-
(A) 100 m
(B) 95 m
(C) 320 m
(D) 160 m​

Answer 1

Answer:

A) 100 m

It is the answer I think so .......

Answer 2

Given : A small ball is released from rest and it accelerates due to gravity,

after 1 sec suddenly gravity disappears and again appears after 5 sec.

Ball takes total 8 sec to reach the ground.

To Find : The height above ground  from where ball was released, is

Solution:

at t= 0  u = 0  as released

S = ut + (1/2)at²

V = u + at

a = g  = 10 m/s²

Distance covered in 1st sec =  0*1 + (1/2)(10)1² = 5 m

Velocity after 1 sec = 0 + 10(1) = 10 m/s

Distance covered in next 4 sec ( 1 to 6 sec )  =  10 * 5 = 50 m

as no acceleration

Hence Velocity after total 6 secs  = 10 m/s²

Distance covered in next 2 secs   ( 6 to 8 sec )

= 10(2) + (1/2)(10)2²

= 20 + 20

= 40m

Total Distance covered =  5 + 50 + 40  = 95 m

height above ground  from where ball was released, is    95 m

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