a body is dropped from a certain height H if the ratio oh the distances travelled by it in (n-3) seconds and (n-3)th second is 4:3 find H (g=10)? answer is 125m

Hello

s_n = u + a(2n-1)/2 ---------------------(1)

s_n is distance covered in nth second

So, distance covered in (n-3)th second = a(2n-7)/2 = 5(2n-7) ---------------(2)

replacing n by (n-3) in equation (1)

here, u=0 and a=10

Also,

distance covered in (n-3) seconds = 1/2 a t² = 5(n-3)² -------------------------(3)

but according to quest.

5(n-3)²/5(2n-7) = 4/3

so, n = 5sec because 11/3 is not possible

So

H = 1/2 a n² = 5*5² = 125metres

plz approve if u like the solution

Anurag Kishore

Approved