# Finding distance -- Find the height of an object given ratio of two distances

D.Man Hazarika

## Homework Statement

A body is dropped from certain height H. If the ratio of the distances travelled by it in (n-3) seconds to (n-3)rd second is 4:3, find H (take g= 10 m/s²) (answer: 125m)

## The Attempt at a Solution

I did this way but I came to a dead end...
To find H,
s= 1/2 gt²
= 1/2 * 10 * (n-3)²
= 5*(n²-9)
= 5n²+45.......(I)

S nth = u+g/2(2n-1)
= 5(2n-1)
= 10n-5.....(II)
equating (I) AND (2)
5n²+45= 4
________ ___

10n-5. 3

=> 3(5n²+45)=4(10n-5)
=> 15n²+135=40n-20
=> 15n²-40n=-20-135
=> 15n²-40n=-155
=> 15n²+155=40n
=> 15n²+155= 40
____________
n

This is a dead end....what should I do next...???

## Answers and Replies

Gold Member
The problem statement looks incomplete. What is n? Is n the number of seconds required to fall the distance H?

Ok, you need to check your algebra a bit. (n-3)^2 is not (n^2 - 9). And 5 (n^2-9) is not 5n^2 + 45. And what are (I) and (II) and why should they be equal?

Also, you need to be using the ratio you have been given. The distance in n-3 seconds is 4/3 the distance in the (n-3)rd second.

• D.Man Hazarika
D.Man Hazarika
Thanks...rechecking my algebra