A heavy particle while falling freely from rest under gravitational altraction , the distance through which it falls varies directly as the square of the time taken to fall up to the distance . The particle falls through 125 metres in 5 seconds . Through what distance will it fall in 10 seconds ? What distance will it traverse in the 10th second?

Explanation

Let distance =h m and the time required to traverse that distance =t second .
As per question , h `prop t^2 or, h = kt^2 ( where k ne 0=` variation constant )
`therefore h =kt^2 ………(1) `
Now if t =5 , then h = 125 , `therefore 125 =k. (5)^2 rArr k =(125)/(25)=5`.
`therefore` from (1) we get , h =`5t^2 ...........(2)`
`therefore` putting t =10 in (2) we get , h = `5xx(10)^2 `
`=5xx100 =500 `.
Again , putting t =9 we get h=`5xx(9)^2 =5xx81 =405` .
Now distance traverse in 10th second
=(Distance travelled in 10 seconds) -(Distance travelled in 9 seconds )
=(500-405) m =95 m .
`therefore` The particle falls 500m in 10 seconds and 95m in 10th seconds.