Question

A ball is dropped from a tall building. The distance between floors in the block is constant. The ball takes 0.5 sec to fall from the 14th to the 13th floor and 0.3 sec to fall from the 13th floor to the 12th. What is the distance between floors?

Answer 1

Distance between two consecutive floors `=0.3 g=2.943\ m`

Explanation:

Let `x` be the distance between two consecutive floors.

The time taken by ball to fall from `14`th to `13`th floor with initial velocity `u` at `14`th floor then the velocity `v` at `13`th floor is given by from equations of motion

`v=u+g t`

`v=u+g(0.5)`

`v=u+0.5g` &

`v^2=u^2+2gh`

`(u+0.5g)^2=u^2+2gx`

`u=\frac{8x-g}{4}\ .........(1)`

Similarly, from `13`th to `12`th floor the initial velocity of ball is `u_1=u+0.5g` at `13`th floor

from equations of motion

`v_1=u_1+g t`

`v_1=u+0.5g+g(0.3)`

`v_1=u+0.8g` &

`v_1^2=u_1^2+2gh`

`(u+0.8g)^2=(u+0.5g)^2+2gx`

`u=\frac{2x-0.39g}{0.6}\ .........(2)`

Now, equating (1) & (2), we get

`\frac{8x-g}{4}=\frac{2x-0.39g}{0.6}`

`3.2x=0.96g`

`x=0.3g`

`=2.943`

hence the distance between two consecutive floors is `2.943\ m`

Answer 2

The distance is `=2.94m`

Explanation:

Let the speed at the beginning of the `14th` floor be `=ums^-1`

Let the distance between the floors be `=hm`

Then applying the equation of motion

`h=ut+1/2g t^2`

The acceleration due to gravity is `g=9.8ms^-2`

Then,

`h=0.5u+4.9*0.5^2=0.5u+1.225`.......................`(1)`

`2h=0.8u+4.9*0.8^2=0.8u+3.136`...................`(2)`

Eliminating `u` from equations `(1)` and `(2)`

`0.5u=h-1.225`

`u=(h-1.225)/0.5`

`2h=0.8*((h-1.225)/0.5)+3.136`

`h=0.8h-0.8*1.225+3.136*0.5`

`0.2h=1.568-0.98=0.588`

`h=0.588/0.2=2.94m`

graph{(y-0.5x-1.225)(2y-0.8x-3.136)=0 [-2.84, 11.204, -0.225, 6.8]}