Question

Physics Challenge , Dare to Solve Question number , 28

Answer 1

let t is the time of free fall of body
velocity of striking time=-gt
velocity of body just 1second before=-g (t-1)
now use v^2=u^2+2as
s=1/2g (1-2t)
now displacement of body first 3second=0-1/2g (3)^2=-45m
a/c question
-45=1/2g (1-2t)
t=5 second
hence time of free fall =5 sec

Answer 2

5 sec is the answer,

first we need to find the displacement for first three second displacement,

= 1/2gt² = 4.905(3)² = 44.145 m Free-Fall displacement during last second = 44.145 m

First three second displacement = 1/2gt² = 4.905(3)² = 44.145 m 

Free-Fall displacement during last second = 44.145 m 

V average during last second of Free-Fall = 44.145/1 = 44.145 m/s 

Vaverage = (Vo+Vf)/2 = 44.145


Vo+Vf = 88.29 
Vo = gt {where t = Free-Fall time - 1 
Vf = 88.29 - 9.81t 

Vf = g(t+1) = 9.81t + 9.81 
88.29 - 9.81t = 9.81t + 9.81 
88.29 - 9.81 = 19.62t 
19.62t = 78.48 
t = 4 s 

time to Free-Fall = t + 1 = 5 s ANS