Question

Can Anyone Solve, Question number 28 ?

Answer 1

First 3 sec of free fall

Displacement = 1/2gt^2
= 4.905(3^2). = 44.451m

free fall of last second
Displacement = 44.451m

V (average) = (v0 + vf)/2

= 44.451m

v0 + vf = 88.29

v0 = gt (t=1)
vf = 88.29 - 9.81 t

Now,
vf = g(t+1)
vf = 9.81t + 9.81

simplify,

by putting value of vf

88.29-9.81t = 9.81t + 9.81
19.62 t = 78.48

thats why,
t = 4,

time to free fall is t + 1,

so 4+1 = 5s.


5s is the answer.

Hope it's helpful to you dear,

All the best

Answer 2

Answer:

Let t is the time of free fall of body

at last second means (t-1) sec velocity of body

v'=0-g (t-1)

v'=-g (t-1) -----(1)

when body strike surface of earth then velocity of body

v=0-gt

v=-gt --------(2)

now displacement in last second

v^2=v'^2+2as

g^2t^2=g^2 (t-1)^2-2gs

-s=(g^2t^2-g^2t^2+2g^2t-g)/2g

=(2gt-g)/2

s=1/2 (1-2t) g

now displacement in first 3second =ut+1/2at^2=0-1/2g (3)^2=-45 m

a/c question

1/2g (1-2t)=-45

1/2g (2t-1)=45

t=5 sec