# A stone is dropped from height h and falls the last half of its distance in 4 secs. Calculate total time and height

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### 1 Answer

Starting its flight from height h, let the stone takes ` t_1` seconds to reach the halfway mark, and `t_2` seconds to reach the ground.

From the equations of motion, the final velocities at those two points are:

`v_1=sqrt(2gh/2)` , i.e. `sqrt(gh)`

and `v_2=sqrt(2gh)`

Time to reach the velocities as mentioned are:

`v_1=0+g*t_1`

`rArr sqrt(gh)=g*t_1`

`rArr t_1=sqrt(gh)/g=sqrt(h/g)`

Similarly,

`v_2=0+g*t_2`

`rArr sqrt(2gh)=g*t_2`

`rArr t_2=sqrt(2gh)/g=sqrt(2h/g)`

By the condition of the problem `t_2-t_1=4s`

`rArr sqrt(2h/g)-sqrt(h/g)=4`

`rArr sqrt(h/g)(sqrt2-1)=4`

`rArr sqrth/g=4/(sqrt2-1)=9.656854`

Squaring both sides,

`h/g=93.25483`

`rArr h=32.2*93.25483=3002.8` ft

and total time of flight, `t_2=sqrt(2h/g)`

`=sqrt(2*93.25483/32.2)=13.66 s`

Check:

`t_1=sqrt(h/g)=sqrt(3002.8/32.2)=9.66 s`

`t_2-t_1=13.66-9.66=4 s`