Question

If a body travels half its total path in the last 1.10 s of its fall from rest, find the total time of its fall (in seconds).

Answer 1

Answer:3.75 s

Step-by-step explanation:

Given Body travels half of its motion in last 1.1 sec

Let h be the height and t be the total time taken

here initial velocity is zero

`h=ut+(gt^2)/(2)`

`h=0+(gt^2)/(2)`

`h=(gt^2)/(2)------1`

Now half of the distance traveled will be in t-1.1 s and half distance traveled is in last 1.1 s

`(h)/(2)=(g\left ( t-1.1\right )^2)/(2)-----2`

from 1 & 2 we get

`gt^2=2g\left ( t-1.1\right )^2`

`t^2-4.4t+2.42=0`

`t=(4.4\pm √(4.4^2-4\left ( 1\right )\left ( 2.42\right )))/(2)`

`t=(4.4\pm 3.11)/(2)`

Therefore two value of t is satisfying the equation but only one value is possible

therefore t=3.75 s