Question

A freely falling body falls through a distance 'h' in the nth
second. In the next second, distance covered by
numerically equal to...
(1) h/g
(2) h+2
(3)h+g
(4) root 2gh + (h/2)

Answer 1

Answer:

thanks for free points ..

Answer 2

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