Question

A body falls from a large height The
ratio of distance traveled travelled in
each time interval to during t=4 to t=3
the journey is

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Answer 1

Step-by-step explanation:

A body falls from height, h=200m. What is the ratio of distance travelled in each 2s, during the=0 to t=6s of the journey?

For fall of a body it starts from initial velocity u = 0 and due to gravitational acceleration ‘g’ it's velocity go on increasing and thus it desends height faster and faster with passage of time.

The governing equations are :

v at time t = u+gt ….(1)

Desend distance, h in time t , h = ut + 1/2 g t^2 ..(2)

v^2 = u^2 + 2 g h ….(3)

So, in first 2s desend distance h02= 1/2 g 2^2 = 2g

The velocity attained at t = 0 to 2s is gt (eqn 2) or 2g

So, desend distance in next 2s, h24 is ut + 1/2 gt^2 or 2g + 1/2 g 4 or 2g + 2g or 4g

The velocity attained at 2+2 or 4th second is gt or 4g and so desend distance from 4s to 6s h46 is

4g *2 + 1/2 g 2^2 or 8g + 2g or 10g

So ratio of desend distance in steps of 2s up to 6s is 2g:4g:10g or 1:2:5

Answer 2

Step-by-step explanation:

C

1:3:5

When dropped then u=0

a=g

Distance travelled in nth sec =S

n

=u+

2

a

(2n−1)

From t=0 to t=1 sec n=1

S

1

=0+

2

g

(2×1−1)=

2

10

=5 m

From t=1 sec to t=2 sec n=2

S

2

=0+

2

g

(2×2−1)=5×3=15 m

From t=2 sec to t=3 sec n=3

S

3

=0+

2

g

(3×2−1)=5×5=25 m

Ratio 5:15:25

=1:3:5