Question

The distance through which a body falls in the nth second
is h. The distance through which it falls in the next second is
(a) h (b)h + g/2
(c) h – g (d) h + g

Answer 1

Answer:

Wkt, according to 2

rd

Kinematic equation ,

S=ut+

2

1

at

2

Distance in n

th

sec=

2

1

gn

2

Distance in (n+1)

th

sec=

2

1

g(n+1)

2

Distance through which it falls in n

th

sec be y

y= Distance in(n+1)

th

sec− Distance in n

th

sec⇒

2

1

g(n+1)

2

−

2

1

gn

2

⇒

2

g

(2n+1)−−−−−(1)

Also, WKT distance in n

th

sec=S

n

=u+

2

a

(2n−1)⇒S

n

=

2

g

(2n−1)−−−−(2)

From (1) and (2)

y=h+g