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

Question

6181 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-13T10:05:57

$please review the details below.$ $"please review the details below."$

$let elapsed time to peak point for faster stone be t_{1}$ $"let elapsed time to peak point for faster stone be " t_1$

$3 v : velocity of faster stone$ $3v:"velocity of faster stone"$

$t_{1} = \frac{3 v}{g}$ $t_1=(3v)/g$

$let elapsed time to peak point for slower stone be t_{2}$ $"let elapsed time to peak point for slower stone be " t_2$

$v : velocity of slower stone$ $v:"velocity of slower stone"$

$t_{2} = \frac{v}{g}$ $t_2=(v)/g$

$Let's find the ratio \frac{t_{1}}{t_{2}}$ $"Let's find the ratio " t_1 / t_2$

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

$t_1/t_2=3$

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

$t_1=10/2=5 sec.$

$5/t_2=3$

$t_2=5/3 sec.$

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

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

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

$"let find the ratio " H/x$

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

$H/x=1/9$

$x=9H$