Question

If the faster stone takes 10 seconds to return to the ground, how long will it take the slower stone to return?

Answer 1

`"please review the details below."`

Explanation:

`"let elapsed time to peak point for faster stone be " t_1`

`3v:"velocity of faster stone"`

`t_1=(3v)/g`

`"let elapsed time to peak point for slower stone be " t_2`

`v:"velocity of slower stone"`

`t_2=(v)/g`

`"Let's find the ratio " t_1 / t_2`

`t_1/t_2=(3cancel(v))/cancel(g)* cancel( g)/cancel(v)`

`t_1/t_2=3`

`"Since the movement is symmetric, the time to reach "`
`"the maximum point is equal to half of the duration of the flight."`

`t_1=10/2=5 sec.`

`5/t_2=3`

`t_2=5/3 sec.`

`"flight time for slower stone "=2*5/3=10/3 sec.`

`H=v^2/(2g)" slower stone's maximum height"`

`x=(3v)^2/(2g)=(9v^2)/(2g)" faster stone's maximum height"`

`"let find the ratio " H/x`

`H/x=cancel(v^2)/(cancel(2g))*(cancel(2g))/(9cancel(v^2)`

`H/x=1/9`

`x=9H`