Question

I want a lot of friendssssss​

Answer 1

Answer:

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

Explanation:

I can be ur friend

Answer 2

Answer:

I also want hehehehehe good night