Question

A stone is dropped from a height such that in 10sit reaches ground.
Find the ratio of distance covered in first and last second of its
motion?

WRITE CORRECTLY FOR BRAINLIEST

Answer 1

Answer:

1 : 19

Explanation:

Using S = ut + 1/2 at^2

Here, u = initial velocity = 0  ( since it is dropped & not thrown)

So, S = 1/2 at^2

Therefore,

 S(1) = 1/2 g(1)^2         {a = g = 9.8}

  S(1) = g/2 m

distance in 10th second = total distance in 10 sec - total distance in 9 sec

 = S(10) - S(9)

 = 1/2 g(10)^2  - 1/2 g(9)^2

 = g/2 (100 - 81) m

 = g/2 (19) m

Required ratio = g/2 : g/2 (19)

                        = 1 : 19

           

Answer 2

Given :-

Height = 10 sit reaches ground

To Find :-

Ratio of 1st and last second

Solution :-

We know that

$\sf \star s = ut + \dfrac{1}{2} a t^{2}$

Let the ratio be S/S'

S/S' =[ 0(1) + 1/2 + g(1)^2]/[1/2 g(10)^2  - 1/2 g(9)^2]

S/S' = [0 + 1/2 + g]/[1/2 + g (100 - 81)]

S/S' = [g/2]/[g/2 (19)]

S/S' = 1:19